NCERT Solutions for Class 11 Maths Chapter 6 - Permutations and Combinations

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Exercise 7.1

1. How many 3 digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that

(i) repetition of the digits is allowed?

(ii) repetition of the digits is not allowed?

Sol. (i) Let the 3-digit number be ABC, when C is at the units place, B at the tens place and A at the hundreds place.

Now, when repetition is allowed, the number of digits possible at C is 5.

As repitition is allowed, the number of digits possible at B and A is also 5 at each.

Hence, the total number possible 3-digit numbers = 5 × 5 × 5 = 125.

(ii) Let the 3-digit numbers be ABC, where C is at the unit’s place, B at the tens place and A at the hundreds place.

Now, when repetition is not allowed the number of digits possible at C is 5.

Let’s suppose, one of 5 digits occupies place C and repitition is not allowed.

So, the possible digits for place B are 4 and similarly there are only 3 possible digits for place A.

Therefore, the total number of possible 3-digit numbers = 5 × 4 × 3 = 60.

2. How many 3 digits even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?

Sol. Let the 3 digit number be ABC, where C is at the unit’s place, B at the tens place and A at the hundreds place.

As the number has to even, the digits possible at C are 2 or 4 or 6. That is number of possible digits at C is 3.

Now, as the repetition is allowed, the digits possible at B is 6.

Similarly, at A also, the number of digits possible is 6.

Therefore, the total number possible 3 digit numbers = 6 × 6 × 3 = 108.

3. How many 4 letters code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?

Sol. Let the 4 digits code be 1234.

At the first place, the number of letters possible is 10.

Suppose any 1 of the ten occupies place 1.

As the repetition is not allowed, the number of letters possible at place 2 is 9.

Now, the number of alphabets left for place 3 is 8 and similarly the number of alphabets possible at 4 is 7.

Therefore the total number of 4 letters code

= 10 × 9 × 8 × 7

= 5040.

4. How many 5 digit telephone numbers can be construted using the digits 0 to 9, if each number starts with 67 and no digit appears more than once?

Sol. Let the five-digit numbers be ABCDE.

Given, first 2 digits of each number is 67. Therefore, the number is 67CDE.

As the repitition is not allowed and 6 and 7 are already taken the digits available for place C are 0, 1, 2, 3, 4, 5, 8, 9.

Therefore, the number of possible digits at place C is 8.

Let one of them is taken at C, now the digits possible at place D is 7 and similarly, at E the possible digits are 6.

Therefore, the total five digit numbers are 8 × 7 × 6 = 336.

5. A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?

Sol. Given a coin is tossed 3 times and the outcomes are recorded.

The possible outcomes after a coin toss are head and tail.

The number of possible outcomes at each coin toss is 2.

Therefore, the total number of possible outcomes after 3 times = 2 × 2 × 2 = 8.

6. Give 5 flags of different colours, how many different signals can be generated, if each signal requires the use of 2 flags, one below the other.

Sol. Given 5 flags of different colours and the signal requires 2 flags.

Therefore, the number of flags possible for upper flag is 5.

Now as one of the flag is taken, the number of flags remaining for lower flag in the signal is 4.

The number of way in which signal can be given = 5 × 4 = 20.

Exercise 7.2

1. Evaluate:

(i) 8 !

(ii) 4 ! – 3 !

Sol. (i) 8 ! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40230.

(ii) 4 ! – 3 ! = (4 × 3 × 2 × 1) – (3 × 2 × 1)

= 24 – 6

= 18.

2. Is 3 ! + 4 ! = 71 ?

Sol. L.H.S. = 3 ! + 4 !

Expanding both factorials

= (3 × 2 × 1) + (4 × 3 × 2 × 1)

= 6 + 24

= 30

R.H.S. = 7 !

Expanding the factorial

= 7 × 6 × 5 × 4 × 3 × 2 × 1

= 5040

Therefore,

L.H.S. ≠ R.H.S.

3 ! + 4 ! ≠ 7 !.

$$\textbf{3. Compute}\space\frac{\textbf{8!}}{\textbf{6!×2!}}.\\\textbf{Sol}\space\frac{8!}{6!×2!}\\\text{Expanding all the factorials}\\\text{ and simplifying, we get}\\=\frac{8×7×6!}{6!×2!}\\=\frac{8×7}{2×1}$$

= 4 × 7

= 28.

$$\textbf{4. If}\space\frac{\textbf{1}}{\textbf{6!}}+\frac{\textbf{1}}{\textbf{7!}}=\frac{\textbf{1}}{\textbf{7!}}\space\textbf{Find x.}\\\textbf{Sol.}\space\text{L.H.S}=\frac{1}{6!}+\frac{1}{7!}\\=\frac{1}{6!}+\frac{1}{7×6!}\\=\frac{7×1+1}{7×6!}\\=\frac{7+1}{7×6!}=\frac{8}{7×6!}\\\text{and R.H.S}=\frac{x}{8!}\space\text{(given)}$$

Equating L.H.S. and R.H.S. we get

$$\frac{8}{7!}=\frac{x}{8!}\\\frac{8}{7!}=\frac{x}{8×7!}\\x=\frac{8×8×7!}{7!}$$

x = 8 × 8

x = 64.

$$\textbf{5. Evaluate}\frac{\textbf{n!}}{\textbf{(n-r)!}},\textbf{when}$$

(i) n = 6, r = 2

(ii) n = 9, r = 5

Sol. (i) Given n = 6 and r = 2

$$\text{Putting the value of n and r in}\space\frac{n!}{(n-r)!},\\\text{we get}\\\frac{6!}{(6-2)!}=\frac{6}{4!}=\frac{6×5×4!}{4!}$$

= 6 × 5 = 30

(ii) Given n = 9 and r = 5

$$\text{Putting the value of n and r in}\\\space\frac{n!}{(n-r)!},\text{we get}\\\frac{9!}{(9-5)!}=\frac{9}{4!}\\=\frac{9×8×7×6×5×4!}{4!}$$

= 9 × 8 × 7 × 6 × 5

= 15120.

Exercise 7.3

1. How many 3 digit numbers can be formed by using the digits 1 to 9, if no digit is repeated?

Sol. Total number of digits possible = 9

Number of places for which a digit has to a be taken = 3

As it is given, no repetition allowed,

∴ No. of permutations

$$=\space^aP_3=\frac{9!}{(9-3)!}\\\bigg[\because\space^nP_r=\frac{n!}{n-r!}\bigg]\\=\frac{9!}{6!}=\frac{9×8×7×6}{6!}$$

= 9 × 8 × 7

= 504.

2. How many 4 digit numbers are there with no digit repeated?

Sol. We need to find four digit number (digits does not repeat)

And, we have 4 places where 4 digits are to be placed.

So, at thousand’s place = There are 9 ways as 0 cannot be at thousand’s place = 9 ways

At hundredth’s place = There are a digits to be filled as 1 digit is already taken = 9 ways

At ten’s place = there are now 8 digits to be filled as 2 digits are already taken = 8 ways

A unit’s place = There are 7 digits that can be filled = 7 ways

Total number of ways to fill the four places = 9 × 9 × 8 × 7 = 4536 ways

So, a total of 4536 four digit numbers can be there with no digits repeated.

3. How many 3 digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?

Sol. Even number means that last digit should be even, means last digit should be divisible by 2.
Numer of possible digit at one’s place = 3(2, 4, 6)

$$\text{Number of permutations =}^3\text{P}_1\\=\frac{3!}{(3-1)!}=\frac{3!}{2!}\\=\frac{3×2!}{2!}=3\space\\\bigg[\space^{n}\text{P}_r=\frac{n!}{(n-r)!}\bigg]$$

One of digits is taken at one’s place,

Number of possible digits available = 5

Number of permutations = ⁵P₂

$$=\frac{5!}{(5-2)!}$$

$$=\frac{5!}{3!}=\frac{5×4×3!}{3!}\\\bigg[\space^n\text{P}_r=\frac{n!}{(n-r)!}\bigg]$$

= 20

Therefore, total no. of permutations = 3 × 20 = 60.

4. Find the number of 4 digit numbers that can be formed using the digits 1, 2, 3, 4, 5, if no digit is repeated. How many of thse will be even?

Sol. Total number of digits possible = 5

Number of places for which a digit has to be taken = 4

As there is no repetition allowed,

$$\text{Number of permutations = }^5\text{P}_4\\\bigg[\space^n\text{P}_r=\frac{n!}{(n-r)!}\bigg]\\=\frac{5!}{(5-4)!}=\frac{5×4!}{1!}\\=\frac{5×4×3×2×1}{1}=120$$

The number will be even, when 2 and 4 are at
one’s place.

The possibility of (2, 4) at one’s place

$$=\frac{2}{5}=0.4$$

Total number of even number = 120 × 0.4 = 48.

5. From a committee of 8 persons, in how many ways can be choose a chairman and a vicechairman assuming one person can not hold more than one position?

Sol. Total number of people in committee = 8

Number of positions to be filled = 2

Number of permutations

$$=\space^8\text{P}_2=\frac{8!}{(8-2)!}\\\bigg[\space^n\text{P}_r=\frac{n!}{(n-r)!}\bigg]\\=\frac{8!}{6!}=\frac{8×7×6}{6!}$$

= 56.

6. Find n, if ^{n – 1}P₃ : ⁿP₄ = 1 : 9.

Sol. n – ¹P₃ : ⁿP₄ = 1 : 9

Given equation,

n – ¹P₃ : ⁿP₄ = 1 : 9

$$\frac{\frac{(n-1)!}{(n-1-3)!}}{\frac{n!}{(n-4)!}}=\frac{1}{9}\\\begin{Bmatrix}\text{using}\space^n\text{P}_r=\frac{n!}{(n-r)!}\end{Bmatrix}\\=\frac{\frac{(n-1)!}{(n-4)!}}{\frac{n!}{(n-4)!}}=\frac{1}{9}\\\text{On simplification}\\\frac{(n-1)!}{n!}=\frac{1}{9}\\\frac{(n-1)!}{n(n-1)!}=\frac{1}{9}\\=\frac{1}{n}=\frac{1}{9}$$

n = 9.

7. Find r if:

(i) ⁵P_r = 2 ⁶Pr – 1

(ii) ⁵P_r = ⁶P_{r – 1.}

Sol. (i) ⁵P_r = 2 ⁶P_{r – 1}

$$\frac{5!}{(5-r)!}=2\frac{6!}{6-(r-1)!}\\\begin{Bmatrix}\text{using\space}^n\text{P}_r=\frac{n!}{(n-r)!}\end{Bmatrix}\\\frac{5!}{(5-r)!}=2\frac{6!}{(7-r)!}\\\frac{7-r!}{5-r!}=2\frac{6!}{5!}\\\text{On simplifying, we get}\\\frac{(7-r)(7-r-1)(7-r-1-1)}{(5-r)!}\\=2×\frac{6×5!}{5!}\\\frac{(7-r)(6-r)(5-r)!}{5-r!}=2×6$$

(7 – r) (6 – r) = 12

42 – 13r + r²= 12

r² – 13r + 30 = 0

r² – 10r – 3r + 30 = 0

{∵ factorization)

r(r – 10) – 3(r – 10) = 0

(r – 3) (r – 10) = 0

r = 3 or r = 10

But r = 10 is rejected as in ⁵Pr r cannot be greater than 5.

Therefore, r = 3

(ii) ⁵P_r = ⁶P_{r – 1}

$$\frac{5!}{(5-r)!}=\frac{6!}{[6-(r-1)]!}\\\begin{Bmatrix}\text{using\space}^n\text{P}_r=\frac{n!}{(n-r)!}\end{Bmatrix}\\\frac{5!}{(5-r)!}=\frac{6!}{(7-r)!}\\=\frac{(7-r)!}{(5-r)!}=\frac{6!}{5!}\\\frac{(7-r)(7-r-1)(7-r-2)!}{(5-r)!}\\=\frac{6×5!}{5!}\\=\frac{(7-r)(6-r)(5-r)!}{(5-r)!}=6$$

(7 – r) (6 – r) = 6

42 – 13r + r² = 6

r² – 13r + 36 = 0

r² – 9r – 4r + 30 = 0

{∵ factorization}

r(r – 9) – 4(r – 9) = 0

(r – 9) (r – 4) = 0

r = 4 or r = 9

But r = 9 is rejected, as in ⁵P_r and cannot be greater than 5.

Therefore, r = 4.

8. How many words, with or without meaning, can be formed using all the letters of the word ‘EQUATION’ using each letter exactly once?

Sol. Total number of different letters in EQUATION = 8

Number of letters to be used to form a word = 8

No. of permutations

= ⁸P₈

$$7=\frac{8!}{8-8!}=\frac{8!}{0!}$$

= 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1

= 40320.

9. How many words, with or without meaning can be made from the letters of the word ‘MONDAY’ assuming that no letter is repeated

if:

(i) 4 letters are used at time.

(ii) all letters are used at a time.

(iii) all letters are used but first letter is a vowel.

Sol. (i) Number of letter to be used = 4

Number of permutations

$$=^6\text{P}_4=\frac{6!}{(6-4)!}\\=\frac{6!}{2!}=\frac{6×5×4×3×2×1}{2×1}$$

= 360.

(ii) Number of letters to be used = 6

No. of permutation

$$=\space^6\text{P}_6=\frac{6!}{(6-6)!}\\=\frac{6!}{0!}=6×5×4×3×2×1$$

= 720.

(iii) Number of vowels in MONDAY = 2(A and O)

Number of permutations in vowel = ²P₁

$$=\frac{2!}{2-1!}=2$$

Now, remaining places = 5

Remaining letters to be used = 5

Number of permutations

$$=\space^5\text{P}_5=\frac{5!}{(5-5)!}\\=\frac{5!}{0!}=5×4×3×2×1$$

= 120

Therefore, total number of permutations

= 2 × 120

= 240.

10. In how many of the distinct permutations of the letters in the ‘MISSISSIPPI’ do the four I’s not come together?

Sol. Total number of letter in MISSISSIPPI = 11

Letter Number of occurence

$$\text{Number of permutations = }\frac{11!}{1!4!4!2!}$$

= 34560

We take that 4I’s come together and they are treated as 1 letter.

∴ Total number of letters = 11 – 4 + 1 = 8

$$\text{Number of permutations =}\frac{8!}{1!4!2!}$$

= 840

Therefore, total number of permutations where four I’s don’t come together

= 34650 – 840

= 33810.

11. In how many ways can the letters of the word ‘PERMUTATIONS’ be arranged, if the:

(i) words start with P and end with S.

(ii) vowels are all together.

(iii) there are always 4 letters between P and S?

Sol. (i) Total number of letters is PERMUTATIONS = 12

∴ Only repeated letter is T = 2 times

First and last letters of the word are fixed as P and S respectively.

Number of letter remaining = 12 – 2 = 10

$$\text{Number of permutations =}\frac{^{10}\text{P}_{10}}{2!}\\=\frac{10!}{(10-10!)2!}\\=\frac{10!}{2}$$

= 1814400.

(ii) Number of vowels in PERMUTATIONS

= 5(E, U, A, I, O)

Now, we consider all the vowels together as
one.

Number of permutations of vowels = 120

Now total number of letters

= 12 – 5 + 1

= 8

Number of permutations

$$=\frac{^8\text{P}_3}{2!}\\=\frac{8!}{2!(8-8)!}=\frac{8!}{2}$$

= 20160

Therefore, total number of permutations

= 120 × 20160

= 2419200.

(iii) Number of places are as

1 2 3 4 5 6 7 8 9 10 11 12

There should always be 4 letters between P and S.

Possible, places of P and S are 1 and 6, 2 and 7, 3 and 8, 4 and 9, 5 and 10, 6 and 11, 7 and 12.

Possible ways = 7

Also, P and S can be interchanged,

Number of permutations = 2 × 7 = 14

Remaining 10 places can be filled with 10

remaining letters,

∴ Number of permutations

$$=\frac{^{10}\text{P}_{10}}{2!}\\=\frac{10!}{2!(10-10)!}=\frac{10!}{2}$$

= 1814400

Therefore, total number of permutations

= 14 × 1814400

= 25401600.

Exercise 7.4

1. If ⁿC₈ = ⁿC₂, find ⁿC_2.

Sol. Given, ⁿC₈= ⁿC₂

We know,

if ⁿC_r = ⁿC_p

then, either r = p or r = n – p

Here ⁿC₈ = ⁿC₂

8 = n –2

n = 10

Now,

∴ ⁿC₂ = ¹⁰C₂

$$=\frac{10!}{2!(10-2)!}\\\bigg[\because\space^n\text{C}_r=\frac{n!}{r!(n-r)!}\bigg]\\=\frac{10×9×8}{2×1×8!}\\=\frac{90}{2}$$

= 45. Ans.

2. Determine n, if:

(i) ²ⁿC₃ : ⁿC₃ = 12 : 1

(ii) ²ⁿC₃ : ⁿC₃ = 11 : 1

Sol. (i) Given,

²ⁿC₃ : ⁿC₃ = 12 : 1

The above equation can be written as

$$\Rarr\space\frac{^{2n}\text{C}_3}{^{n}\text{C}_3}=\frac{12}{1}\\\text{Substituting the formula}\\=\frac{\frac{2n!}{3!(2n-3)!}}{\frac{n!}{3!(n-3)!}}\\\bigg[\space^{n}\text{C}_r=\frac{n!}{r!(n-r)!}\bigg]\\=\frac{12}{1}\\=\frac{\frac{2n!}{(2n-3)!}}{\frac{n!}{(n-3)!}}=\frac{12}{1}$$

Expanding the factorial, we get

$$\Rarr\space\frac{\frac{2(n)×(2n-1)(2n-2)(2n-3)!}{(2n-3)!}}{\frac{n(n-1)(n-2)(n-3)!}{(n-3)!}}=\frac{12}{1}\\\text{On simplifying}$$

$$\Rarr\space\frac{2n×(2n-1)(2n-2)}{n(n-1)(n-2)}=\frac{12}{1}\\\Rarr\space\frac{2(2n-1)2(n-1)}{(n-1)(n-2)}=\frac{12}{1}\\\Rarr\space\frac{4(2n-1)}{(n-2)}=\frac{12}{1}$$

⇒ 4(2n – 1) = 12(n – 2)

{·.· By cross multiplication}

⇒ 8n – 4 = 12n – 24

⇒ 12n – 8n = 24 – 4

⇒ 4n = 20

⇒ n = 5

Hence, the value of n is 5.

(ii) Given, ²ⁿC₃ : ⁿC₃ = 11 : 1

The above equation can be written as

$$\frac{^{2n}\text{C}_3}{^n\text{C}_3}=\frac{11}{1}\\\text{Substituting the formula}\space\\\ ^{n}\text{C}_r=\frac{n!}{r!(n-r)!}\\\frac{\frac{2n!}{3!(2n-3)!}}{\frac{n!}{3!(n-3)!}}=\frac{11}{1}\\\text{Expanding the factorial}\\\frac{\frac{2n(2n-1)(2n-2)(2n-3)!}{3!(2n-3)!}}{\frac{n(n-1)(n-2)(n-3)!}{3!(n-3)!}}=\frac{11}{1}$$

On simplifying,

$$\frac{2n(2n-1)(2n-2)}{n(n-1)(n-2)}=\frac{11}{1}\\\frac{2n(2n-1)2(n-1)}{n(n-1)(n-2)}=\frac{11}{1}\\\frac{4(2n-1)}{(n-2)}=\frac{11}{1}$$

4(2n – 1) = 11(n – 2)

{·.· By cross multplication}

8n – 4 = 11n – 22

11n – 8n = 22 – 4

3n = 18

n = 6

Hence, the value of n is 6.

3. How many chords can be drawn through 21 points on a circle?

Sol. Given 21 points on a circle.

As we know that we require two points on the circle to draw a chord.

∴ Number of chord are = ²¹C₂

$$^{21}\text{C}_{2}=\frac{21!}{2!(21-2)!}\\\begin{Bmatrix}\because\space ^n\text{C}_r=\frac{n!}{r!(n-r)!}\end{Bmatrix}\\=\frac{21×20×19!}{2!×19!}\\=\frac{21×20}{2×1}\\=\frac{420}{2}$$

= 210

⇒ Total number of chords can be drawn are 210.

4. In how many way can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?

Sol. Given 5 boys and 4 girsl are in total we can select 3 boys from 5 boys in ⁵C₃ ways.

Similarly, we can select 3 girls from 4 girls in ⁴C₃
ways.

∴ Number of ways a team of 3 boys and 3 girls can be selected in ⁵C₃ × ⁴C₃.

$$\Rarr\space^5\text{C}_3×^4\text{C}_3=\\\frac{5!}{3!(5-3)!}×\frac{4!}{3!(4-3)!}\\=\frac{5!}{3!2!}×\frac{4!}{3!1!}\\=\frac{5×4×3!}{3!×2}×\frac{4×3!}{3!×1}$$

= 5 × 4 × 2

= 10 × 4

= 40

Hence, number of ways are 40.

5. Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.

Sol. Given: Red balls = 6

White balls = 5

Blue balls = 5

We can select 3 red balls from 6 red balls in ⁶C₃ways.

Similarly, we can select 3 white balls from 5 white balls in ⁵C₃ ways.

Again, we can select 3 blue balls from 5 blue balls in ⁵C₃ ways.

∴ Number of ways of selecting 9 balls are ⁶C₃× ⁵C₃ × ⁵C₃

⁶C₃ × ⁵C₃ × ⁵C₃

$$=\frac{6!}{3!(6-3)!}×\frac{5!}{3!(5-3)!}×\frac{5!}{3!(5-3)!}\\=\frac{6!}{3!3!}×\frac{5!}{3!2!}×\frac{5!}{3!×2!}\\=\frac{6×5×4×3!}{3!×3!}×\frac{5×4×3!}{3!×2!}\\×\frac{5×4×3!}{3!×2!}\\\frac{6×5×4}{3×2}×\frac{5×4}{2}×\frac{5×4}{2}$$

= 20 × 10 × 10

= 2000

Hence, there are 2000 ways to select 9 balls.

6. Determine the number of 5 card combinations out of a deck of 52 cards, if there is exactly one ace in each combination.

Sol. Given, a deck of 52 cards.

There are 4 ace cards in a deck of 52 cards.

According to question,

We need to select 1 Ace card out of 4 Ace cards.

∴ Number of ways to select 1 Ace from 4 Ace card is ⁴C_1.

More 4 cards are to be selected now from 48 cards (52 cards – 4 Ace cards)

∴ Number of ways to select 4 cards from 48 cards is ⁴⁸C_4.

$$\Rarr\space ^4\text{C}_1×^{48}\text{C}_4\\=\frac{4!}{1!(4-1)!}×\frac{48!}{4!(48-4)!}\\=\frac{4!}{1!3!}×\frac{48!}{4!×44!}\\=\frac{4×3!}{1×3!}×\frac{48×47×46×45×44!}{4!44!}\\=4×\frac{4669920}{9×3×2×1}\\=778320.$$

7. In how many ways can one select a cricket team of elevan from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?

Sol. Given: 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers.

There are 5 players who bowl, and we can require 4 bowlers in a team of 11.

∴ Number of ways in which bowlers can be selected are ⁵C_4.

Now other players, left are = 17 – 5 (bowlers) = 12

Since, we need 11 players in a team and already 4 bowlers are selected, we need to select 7 more players from 12.

∴ Number of ways we can select these players are ¹²C_7.

∴ Total number of combinations possible are ⁵C₄ × ¹²C₇

$$^5\text{C}_4×^{12}\text{C}_7=\\\frac{5!}{4!(5-4)!}×\frac{12!}{7!(12-7)!}\\=\frac{5!}{4!1!}×\frac{12!}{7!5!}\\=\frac{5×4!}{1×4!}×\frac{12×11×10×9×8×7!}{5×4×3×2×1×7!}$$

= 3960. Ans.

8. A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.

Sol. Given, a bag contains 5 black and 6 red balls.

Number of ways we can select 2 black balls from 5 black balls are ⁵C_2.

Number of ways we can select 3 red balls from 6 red balls are ⁶C_3.

Number of ways 2 black and 3 red balls can be selected are ⁵C₂ × ⁶C_3.

$$\therefore\space^5\text{C}_2×^6\text{C}_3=\frac{5!}{2!(5-2)!}×\frac{6!}{3!(6-3)!}\\=\frac{5!}{2!3!}×\frac{6!}{3!×3!}\\=\frac{5×4×3!}{2×1×3!}×\frac{6×5×4×3!}{3!×3×2×1}$$

= 5 × 2 × 5 × 4

= 10 × 20

= 200

Hence, number of ways in which 2 black balls and 3 red balls can be selected from 5 black and 6 red balls are 200. Ans

9. In how many ways can a student choose a program of 5 courses, if 9 courses are available and 2 specific courses are compulsory for every
student?

Sol. Given, 9 courses, are available and 2 specific courses are compulsory for every student.

Here 2 courses are compulsory out of 9 courses.

So, a student need to select 5 – 2 = 3 courses.

∴ Number of ways in which 3 ways can be selected from 9 – 2 (compulsory courses) = 7 are ⁷C₃

$$\therefore\space^7\text{C}_3=\frac{7!}{3!(7-3)!}=\frac{7!}{3!4!}\\=\frac{7×6×5×4!}{3×2×4!}$$

= 7 × 5

= 35

Hence, number of ways a student select 5 courses from 9 courses where 2 specific courses are compulsory are 35.

Miscellaneous Exercise

1. How many words, with or without meaning each of 2 vowels and 3 consonants can be formed from the letters of the word ‘DAUGHTER’?

Sol. The word ‘DAUGHTER’ has 3 vowels A, E, U and 5 consonants D, G, H, T and R.

The two vowels can be chosen in 3C2 as only two vowels are to be chosen.

Similarly, the three consonants can be chosen in ⁵C₃ ways.

∴ Number of choosing 2 vowels and 5 consonants would be ³C₂ × ⁵C₃

$$=\frac{3!}{2!(3-2)!}×\frac{5!}{3!(5-3)!}\\=\frac{3!}{2!1!}×\frac{5!}{3!2!}\\=\frac{3×2!}{2!}×\frac{5×4×3!}{3!×2!}\\=\frac{3×5×4}{2}$$

= 3 × 10 = 30

The number of ways of is 30.

Each of these 5 letters can be arranged in 5 ways to form different words = ⁵P₅

$$=\frac{5!}{(5-5)!}=\frac{5!}{0}=\frac{5!}{1}$$

= 5 × 4 × 3 × 2 × 1 = 120

Total number of words formed would be

= 30 × 120 = 3600. Ans.

2. How many words, with or without meaning, can be formed using all the letters of the word ‘EQUATION’ at a time so that the vowels and consonants occur together?

Sol. In the word EQUATION there are 5 vowels (A, E, I, O, U) are 3 consonants (Q, T, N).

The number of ways in which 5 vowels can be arranged are ⁵C₅

$$=\frac{5!}{(5-5)!}=\frac{5×4×3×2×1}{0!}$$

= 120 …(i)

Similarly, the numbers of ways in which 3 consonants can be arranged are 3P3

$$\Rarr\space\frac{3!}{(3-3)!}=\frac{3×2×1}{0!}$$

= 6 …(ii)

There are two ways in which vowels and consonants are appear together.

(AEIOU) (QTN) or (QTN) (AEIOU)

∴ The total number of ways in which vowel and consonant can appear together are 2 × ⁵C₅ × ³C₃

∴ 2 × 120 × 6 = 1440. Ans

3. A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of:

(i) exactly 3 girls?

(ii) atleast 3 girls?

(iii) atmost 3 girls?

Sol. (i) Given, exactly 3 girls.

Total number of girls are 4.

Out of which 3 are to be chosen.

∴ Number of ways in which choice would be made = ⁴C₃

Number of boys are 9 out of which 4 are to be chosen which is given by ⁹C_4.

Total ways fo forming the committee with exactly three girls = ⁴C₃ × ⁹C₄

$$=\frac{4!}{3!(4-3)!}×\frac{9!}{4!(9-4)!}\\\begin{Bmatrix}\because\space ^n\text{C}_r=\frac{n!}{(n-r)!}\end{Bmatrix}\\=\frac{4!}{3!1!}×\frac{9!}{4!5!}\\=\frac{9×8×7×6×5×4×3×2×1}{3×2×1×5×4×3×2×1}\\=504\space\textbf{Ans}.$$

(ii) Given at least 3 girls.

There are two possibilities of making committee choosing at least 3 girls.

There are 3 girls and 4 boys or there are 4 girls and 3 boys.

Choosing three girls we have done in (i) choosing four girls would be done in ⁴C₄ways.

And choosing 3 boys would be done in ⁹C₃

Total ways = ⁴C₄ × ⁹C₃

$$=\frac{4!}{4!(4-4)!}×\frac{9!}{3!(9-3)!}\\=\frac{4!}{4!0!}×\frac{9!}{3!6!}\\=\frac{9×8×7×6!}{3×2×1×6!}$$

= 84

Total number of ways of making the committes are 504 + 84 = 588. Ans.

(iii) Given at most 3 girls.

In this case the numbers of possibilities are

0 girls and 7 boys

1 girl and 6 boys

2 girls and 5 boys

3 girls and 4 boys

Number of ways to choose 0 girl and 7 boys

$$=\space^4\text{C}_0×^9\text{C}_7\\=\frac{4!}{0!(4-0)!}×\frac{9!}{7!2!}\\=\frac{4!}{4!}×\frac{9×8×7!}{7!×2×1}\\=\frac{72}{2}=36$$

Number of ways of choosing 1 girl and 6 boys

= ⁴C₁ × ⁹C₆

$$=\frac{4!}{1!3!}×\frac{9!}{6!3!}\\=\frac{4×3!}{3!}×\frac{9×8×7×6!}{6!×3×2×1}$$

= 336

Number of ways of choosing 2 girls and 5 boys

$$=^4\text{C}_2×^9\text{C}_5\\\frac{4!}{2!(4-2)!}×\frac{9!}{5!(9-5)!}\\=\frac{4!}{2!2!}×\frac{9!}{5!4!}\\=\frac{4×3×2!}{2!×2×1}×\frac{9×8×7×6×5!}{5!×4×3×2×1}\\=756$$

Number of choosing 3 girls and 4 boys has been done in (i) = 504.

Total number of ways in which comitee can have at most 3 girls are

= 36 + 336 + 756 + 504

= 1632. Ans.

4. If the different permutations of all the letters of the word ‘EXAMINATION’ are listed as in a dictionary, how many words are there in this list before the first word starting with E?

Sol. In dictionary words are listed alphabetically, so to find the words.

Listed before E should start with letter either A, B, C or D.

But the word EXAMINATION doesn’t have B, C and D.

Hence the words should start with letter A.

The remaining 10 places are to be filled by the remaining letters fo the word EXAMINATION which are E, X, A M, 2N, T, 2I, O.

Since, the letters are repeating the formula used would be

$$=\frac{n!}{\text{P}_1!\text{P}_2!\text{P}_3!}$$

where n is remaining number of letter P₁ and P₂are number of times the repeated terms occurs

$$=\frac{10!}{2!2!}$$

= 907200

The number of words in the listed before the word starting with E = words starting with letter A = 907200. Ans.

5. How many 6 digits numbers can be formed from the digts 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated?

Sol. The numbers is divisible by 10 if the unit place has 0 in it.

The 6 digit number is to be formed out of which unit place is fixed as 0.

The remaining to places can be filled by 1, 3, 5, 7 and 9.

Here n = 5

And the numbers of choice available are 5.

So the total ways in which the rest the places can be filled are ⁵P₅

$$=\frac{5!}{(5-5)!}$$

= 5 !

= 5 × 4 × 3 × 2 × 1 × 1

= 120. Ans.

6. The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet?

Sol. We know that there are 5 vowels and 21 consonants in English alphabets.

Choosing two vowels out of 5 would be done in ⁵C₂ ways.

Choosing two consonants out of 21 can be done in ²¹C₂ ways.

The total number of ways selecting 2 vowels and 2 consonants.

= ⁵C₂ × ²¹C₂

$$=\frac{5!}{2!2!}×\frac{21!}{21!9!}\\=\frac{5×4×3!}{2!3!}×\frac{21×20×11!}{2×1×19!}$$

= 2100

Each of these four letters can be arranged in four ways ⁴P₄

$$=\frac{4!}{0!}$$

= 4 × 3 × 2 × 1 = 24 ways

Total numbers of words that can be formed are 24 × 2100 = 50400. Ans.

7. In an examination, a question paper consists of 12 questions divided into two parts i.e., part I and part II, containing 5 and 7 questions, respectively. A student is required to attempt 8
questions in all selecting at least 3 from each part. In how many ways can a student select the questions.

Sol. The student can choose 3 questions, from Part I and 5 from Part II. or 4 questions from part I and 4 from part II. or 5 question from part I and 3 from part II. 3 questions from part 1 and 5 from part II can be chosen in = ⁵C₃ × ⁷C₅

$$=\frac{5!}{3!(5-3)!}×\frac{7!}{5!(7-5)!}\\=\frac{5×4×3!}{3!×2×1}×\frac{7×6×5!}{5!×2×1}$$

= 210

4 questions from part I and 4 from part II can be chosen in = ⁵C₄ × ⁷C₄

$$=\frac{5!}{4!(5-4)!}×\frac{7!}{4!×(7-4)!}\\=\frac{5×4!}{4!}×\frac{7!}{4!3!}\\=5×\frac{7×5×5×4!}{4!×3×2×1}$$

= 175

5 questions from part I and 3 from part II can be chosen in ⁵C₅ × ⁷C₃

$$=\frac{5}{5!0!}×\frac{7!}{3!4!}\\=1×\frac{7×6×5×4!}{3×2×1×4!}$$

= 35

Now the total number of ways in which a student can choose the questions are

= 210 + 175 + 35

= 420.

8. Determine the number of 5 card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.

Sol. We have a deck of card which has 4 kings.

The ways of selecting a king from the deck = ⁴C₁ numbers of cards are 52.

Ways of selecting the remaining 4 cards from 48 cards = ⁴⁸C₄

Total number of selecting the 5 cards having one king always = ⁴C₁ × ⁴⁸C₄

$$=\frac{4!}{1!(4-1)!}×\frac{48!}{4!(48-4)!}\\=\frac{4×3!}{3!}×\frac{48×47×46×45×44!}{4×3×2×44!}$$

= 778320. Ans.

9. It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?

Sol. Given there are total 9 people.

Women occupies even places that means they will be sitting 2^nd, 4^th, 6^th and 8^th place where as men will be sitting one 1^st, 3^rd, 5^th, 7^th and 9^th place.

4 women can sit in four places and ways they can be stated = ⁴P₄

$$=\frac{4!}{(4-4)!}=\frac{4×3×2×1}{(0!)}$$

= 24

5 men can occupy 5 seats in ways.

The number of ways in which there can be seated

$$=\space^5\text{P}_5=\frac{5!}{(5-5)!}\\=\frac{5×4×3×2×1}{0!}=120$$

The total numbers of sitting arrangements possible are 24 × 120 = 2880. Ans.

10. From a class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can the excursion party be chosen?

Sol. In this question we get 2 options.

(i) Either all 3 will go.

Then, remaining students in class are

= 25 – 3

= 22

Number of students remained to be chosen for party = 7

Number of ways choosing the remaining 22

$$\text{students}=\space^{22}\text{C}_7=\frac{22!}{7!(22-5)!}\\=\frac{22!}{7!5!}\\=\frac{22×21×20×19×18×17×16×15!}{15!×7×6×5×4×3×2×1}$$

= 170544.

(ii) None of them will go the students going will be 10.

Remaining students eligible for going = 22

Number of ways in which these 10 students can be selected are