Chapter 6: Permutations and Combinations

Ans:  There are 3 vowels i.e. A, U and E and 5 consonants i.e. D, G, H, T and R in the given word.

Therefore, the number of ways of selecting 2 vowels out of 3 = $^{3}{{C}_{2}}$=3

Number of ways of selecting 3 consonants out of 5 = $^{5}{{C}_{3}}=10$

Thus, by multiplication principle, the number of combinations = $3\times 10=30$.

Each of these combinations can be arranged in 5! ways.

Hence, the number of different words = $30\times 5!=3600$.

Ans: There are 5 vowels i.e. A, E, I, O and U and 3 consonants i.e. Q, T and N.

Since, vowels and consonants occur together, both (AEIOU) and (QTN) can be considered as single objects.

Thus, there are 5! Permutations of 5 vowels taken all at a time and 3! permutations of 3 consonants taken all at a time.

Therefore, by multiplication principle, the number of words = $2!\, \times \,5!\, \times \,3!\, = \,1440$.

Ans: (i) Out of 9 boys and 4 girls, a committee of 7 has to be formed.

Given: exactly 3 girls should be there in a committee, hence, the number of boys = (7 - 3) = 4 boys only.

Therefore, many ways 

$^{4}{{C}_{3}}{{\times }^{9}}{{C}_{4}}=\frac{4!}{3!1!}\times \frac{9!}{4!5!}$

= $4\times \frac{9\times 8\times 7\times 6\times 5!}{4\times 3\times 2\times 5!}$

= 504

(ii) Given: at least 3 girls are required in each committee. This can be done in 2 ways

(a) 3 girls and 4 boys or (b) 4 girls and 3 boys

3 girls and 4 boys can be selected in  $^{4}{{C}_{3}}{{\times }^{9}}{{C}_{4}}$ ways.

4 girls and 3 boys an be selected in  $^{4}{{C}_{4}}{{\times }^{9}}{{C}_{3}}$ ways.

Thus, the number of ways =

$^{4}{{C}_{3}}{{\times }^{9}}{{C}_{4}}{{+}^{4}}{{C}_{4}}{{\times }^{9}}{{C}_{3}}$

= 504+84

= 588

(iii) Given: at most 3 girls in every committee. This can be done in 4 ways

(a) 3 girls and 4 boys (b) 2 girls and 5 boys.

(c) 1 girl and 6 boys (d) No girl and 7 boys

3 girls and 4 boys can be selected in $^{4}{{C}_{3}}{{\times }^{9}}{{C}_{4}}$ ways

2 girls and 5 boys can be selected in $^{4}{{C}_{2}}{{\times }^{9}}{{C}_{5}}$ ways

1 girl and 6 boys can be selected in  $^{4}{{C}_{1}}{{\times }^{9}}{{C}_{6}}$ ways

No girl and 7 boys can be selected in $^{4}{{C}_{0}}{{\times }^{9}}{{C}_{7}}$ ways.

Thus, many ways

$ {{=}^{4}}{{C}_{3}}{{\times }^{9}}{{C}_{4}}{{+}^{4}}{{C}_{2}}{{\times }^{9}}{{C}_{5}}{{+}^{4}}{{C}_{1}}{{\times }^{9}}{{C}_{6}}{{+}^{4}}{{C}_{0}}{{\times }^{9}}{{C}_{7}} $

 $ =\frac{4!}{3!1!}\times \frac{9!}{4!5!}+\frac{4!}{2!2!}\times \frac{9!}{5!4!}+\frac{4!}{1!3!}\times \frac{9!}{6!3!}+\frac{4!}{0!4!}\times \frac{9!}{7!2!} $ 

 $ =504+756+336+36 $ 

$=1632 $

Ans: There are a total of 11 letters out of which A, I and N appear 2 times and other letters appear only once.

The words starting with A will be the words listed before the words starting with E.

Thus, words starting with the letter A will have the letter A fixed at its extreme left end.

And remaining 10 letters are rearranged all at a time.

In the remaining 10 letters, there are 2 I’s and 2 N’s.

Number of words starting with A = $\dfrac{{10!}}{{2!2!}}\, = \,907200$

Therefore, the required number of words is 907200.

Ans: A number can be divisible by 10 only if its unit digit is 0.

Hence, 0 is fixed at the unit place.

Therefore, the 5 vacant places can be filled by the remaining 5 digits.

These 5 empty places can be filled in 5! ways.

Therefore, number of 6-digit numbers = 5! = 120.

Ans: Given: 2 vowels and 2 consonants should be selected from the English alphabet.

We know that there are 5 vowels.

Thus, the number of ways of selecting 2 vowels out of 5 = \[^{5}{{C}_{2}}=\frac{5!}{2!3!}=10\]

Also, we know that there are 21 consonants.

Thus, the number of ways of selecting 2 consonants out of 21 = \[^{21}{{C}_{2}}=\frac{21!}{2!19!}=210\]

Thus, the total number of combinations of selecting 2 vowels and 2 consonants is

= $10\, \times \,210\, = \,2100$.

Every 2100 combination consists of 4 letters, which can be arranged in 4! ways.

Therefore, number of words = $2100\, \times \,4!\, = \,50400$

Ans:  Given: 12 Qs are divided into 2 parts – Part I and Part II consisting of 5 and 7 Qs respectively. A student must attempt 8 Qs with at least 3 from each part. This can be done as:

1. 3 Qs from part I and 5 Qs from part II.

2. 4 Qs from part I and 4 Qs from part II.

3. 5 Qs from part I and 3 Qs from part II.

The first case can be selected in $^{5}{{C}_{3}}{{\times }^{7}}{{C}_{5}}$ ways.

The second case can be selected in $^{5}{{C}_{4}}{{\times }^{7}}{{C}_{4}}$ ways.

The third case can be selected in $^{5}{{C}_{5}}{{\times }^{7}}{{C}_{3}}$ ways.

Thus, number of ways of selecting Q’s

${{=}^{5}}{{C}_{3}}{{\times }^{7}}{{C}_{5}}{{+}^{5}}{{C}_{4}}{{\times }^{7}}{{C}_{4}}{{+}^{5}}{{C}_{5}}{{\times }^{7}}{{C}_{3}}$ 

 $ =\frac{5!}{2!3!}\times \frac{7!}{2!5!}+\frac{5!}{4!1!}\times \frac{7!}{4!3!}+\frac{5!}{5!0!}\times \frac{7!}{3!4!} $

$ =210+175+35 $

$=420 $

Ans:  Out of 52 cards, 5-card combinations have to be made in such a way that in each selection of 5 cards, there is exactly one king. We know that in 52 cards, there are 4 kings.

Out of 4 kings, 1 king can be selected in $^{4}{{C}_{1}}$ ways.

Out of the remaining 48 cards, 4 cards can be selected in $^{48}{{C}_{4}}$ ways.

Therefore, number of 5-card combinations =  $^{4}{{C}_{1}}{{\times }^{48}}{{C}_{4}}$ ways.

Ans: Given 5 men and 4 women should be seated so that women always occupy the even places.

Thus, the men can be seated in 5! ways.

For each arrangement, the women can be seated only in even places.

Thus, the women can be seated in 4! ways.

Therefore, possible number of arrangements = $4!\, \times \,5!\, = \,24\, \times \,120\, = \,2880$

Ans:  Given: 10 are chosen for an excursion party out of 25 students.

There are 2 cases where 3 students decided either all or one of them would join.

Case 1: All the 3 students join.

The remaining 7 students can be chosen out of 22 students in $^{22}{{C}_{7}}$ ways.

Case 2: None of the 3 students join.

Thus, 10 students can be chosen in $^{22}{{C}_{10}}$ ways.

Hence, number of ways of choosing excursion party = $^{22}{{C}_{7}}{{+}^{22}}{{C}_{10}}$