Permutation and combination-Exercise 1.3

1.Please see the question from book.

Soln:

3 marbles can be chosen from 10 marbles in:

C(10,3) = 10!7!.3!$\frac{10!}{7!.3!}$ = 10∗9∗83∗2∗1$\frac{10\ast 9\ast 8}{3\ast 2\ast 1}$ = 120 ways.

2.Please see the question from book.

Soln:

The student can be select 5 courses out of 8 courses in

C(8,5) = 8!(8−5)!5!$\frac{8!}{\left(8-5\right)!5!}$ = 8∗7∗6∗5!3∗2∗1∗5!$\frac{8\ast 7\ast 6\ast 5!}{3\ast 2\ast 1\ast 5!}$ = 56 ways.

Out of 8 courses 3 courses are compulsory, therefore, there is selection = 5 – 3 = 2 courses out if 8 – 3 = 5 courses is C (5,2).

= 5!(5−2)!.12!$\frac{5!}{\left(5-2\right)!.12!}$ = 5∗4∗3∗2∗13∗2∗1∗2∗1$\frac{5\ast 4\ast 3\ast 2\ast 1}{3\ast 2\ast 1\ast 2\ast 1}$ = 10 ways.

3.Please see the question from book.

Soln:

(i) When one particular person is always included, 3 persons have to be selected from remaining 9 persons. This can be selected in:

C(9,3) = 9!6!.3!$\frac{9!}{6!.3!}$ = 9∗8∗73∗2∗1$\frac{9\ast 8\ast 7}{3\ast 2\ast 1}$ = 84 ways.

(ii) When two particular persons are always excluded, 4 persons have to be selected from remaining 8 persons. This can be selected in:

C(8.4) = 8!4!.4!$\frac{8!}{4!.4!}$ = 8∗7∗6∗54∗3∗2$\frac{8\ast 7\ast 6\ast 5}{4\ast 3\ast 2}$ = 70 ways.

4.Please see the question from book.

Soln:

5 white balls can be drawn from 8 white balls in C(8,5) different ways.

3 blue balls can be drawn from 5 blue balls in C(5,3) different ways,

Therefore, total no. of selection (drawn) = C(8,5) * C(5,3)

= 8!(8−5)!.5!$\frac{8!}{\left(8-5\right)!.5!}$ * 5!(5−3)!.3!$\frac{5!}{\left(5-3\right)!.3!}$.

= 56 * 10 = 569.

5.Please see the question from book.

Soln:

4 boys can be chosen from 7 boys in C(7,4) different ways.

3 girls can be chosen from 5 girls in C(5,3) different ways.

= 7!(7−4)!−4!$\frac{7!}{\left(7-4\right)!-4!}$ * 5!(5−3)!.3!$\frac{5!}{\left(5-3\right)!.3!}$.

= 35 * 10 = 350.

6.Please see the question from book.

Soln:

3 men can be chosen from 11 men is

Or, C(11,3) = 11!8!.3!$\frac{11!}{8!.3!}$ = 11∗10∗93∗2∗1$\frac{11\ast 10\ast 9}{3\ast 2\ast 1}$ = 165 ways.

2 women can be chosen from 8 women in

C(8,2) = 8!6!.2!$\frac{8!}{6!.2!}$ = 8∗72∗1$\frac{8\ast 7}{2\ast 1}$ = 28 ways.

So, total no. of committee = 165 * 28 = 4,620.

7.Please see the question from book.

Soln:

2 members can be selected from 4 mathematician in C(4,2) different ways.

2 members can be selected from 6 statistician in C(6,2) different ways.

2 members can be selected from 5 economics in C(5,2) different ways.

Therefore, total number of committees = C(4,2) * C(6,2) * C(5,2) = 6 * 15 * 10 = 900.

8.Please see the question from book.

Soln:

No. or relatives = 8

No, of other = 12 – 8 = 4

5 relatives can be selected from 8 in

C(8,5) = 8!5!.3!$\frac{8!}{5!.3!}$= 8.7.63.2.1$\frac{\mathrm{8.7.6}}{\mathrm{3.2.1}}$ = 56 ways.

2 others can be selected from 4 in

C(4,2) = 4!2!.2!$\frac{4!}{2!.2!}$ = 4.32.1$\frac{4.3}{2.1}$ = 6 ways.

So, total no. of selection = 56 * 6 = 336 ways.

9.Please see the question from book.

Soln:

Defective bulbs = 3,

Non – defective (good) bulbs = 10 – 3 = 7.

When the selection of 6 bulbs is made so that 4 of them may be good then there is selection of 2 bulbs from defective bulbs.

Now, the selection of 2 bulbs from 3 defective bulbs in C(3,2) different ways.

The selection of 4 bulbs from 7 good bulbs = C(7,4) different ways.

Therefore, total number of selection = C(3,2) * C(7,4)

= 3!(3−2)!.2!$\frac{3!}{\left(3-2\right)!.2!}$ * 7!(7−4)!.4!$\frac{7!}{\left(7-4\right)!.4!}$

= 3 * 35 = 105.

10.Please see the question from book.

Soln:

Ladies	Gentlemen	Selection
1 2 3 4	4 3 2 1	C(4,1) * C(6,4) C(4,2) * C(6,3) C(4,3) * C(6,2) C(4,4) * C(6,1)

Total no. of selection = C(4,1) * C(6,4) + C(4,2) * C(6,3) + C(4,3) * C(6,2) + C(4,4) * C(6,1)

= 4 * 6∗52$\frac{6\ast 5}{2}$ + 4∗32$\frac{4\ast 3}{2}$ * 6∗5∗43∗2∗1$\frac{6\ast 5\ast 4}{3\ast 2\ast 1}$ + 4 * 6∗52$\frac{6\ast 5}{2}$ + 1 * 6.

= 60 + 120 + 60 + 6 = 246.

11.Please see the question from book.

Soln:

1st Group	2nd group	Selection
4 3 2	2 3 4	C(5,4) * C(5,2) C(5,3) * C(5,3) C(5,2) * C(5,4)

Total no. of selections = C(5,4) * C(5,2) + C(5,3) * C(5,3) + C(5,2) * C(5,4)

= 5 * 5∗42$\frac{5\ast 4}{2}$ + 5∗42$\frac{5\ast 4}{2}$ * 5+42$\frac{5+4}{2}$ + 5∗42$\frac{5\ast 4}{2}$ * 5.

= 50 + 100 + 50 = 200.

12.Please see the question from book.

Soln:

A man can invite either in 1 or 2 or 3 or 4 or 5 to his dinner.
Therefore, total number of invitation.

= C(5,1) + C(5,2) + C(5,3) + C(5,4) + C(5,5)

= 5!(5−1)!.1!$\frac{5!}{\left(5-1\right)!.1!}$ + 5!(5−2)!.2!$\frac{5!}{\left(5-2\right)!.2!}$ + 5!(5−3)!.3!$\frac{5!}{\left(5-3\right)!.3!}$ + 5!(5−4)!.4!$\frac{5!}{\left(5-4\right)!.4!}$ + 5!(5−5)!.5!$\frac{5!}{\left(5-5\right)!.5!}$

= 5 + 10 + 10 + 5 + 1 = 31.

13.Please see the question from book.

Soln:

a.

Given C(20,r + 5) = C(20,2r – 7)

Then r + 5 = 2r – 7      [if C(n,r) = C(n,r’) then r = r’]

Or, r = 12.

So, C(15,r) = C(15,12) = 15!(15−12)!.12!$\frac{15!}{\left(15-12\right)!.12!}$ = 455.

b.

Given. C(n,10) + C(n,9) = C(20,10)

Or, C(n + 1,10) = C(20,10)     [if C(n,r)+ C(n,r – 1) = C(n + 1, r)]

So, n + 1 = 20

So, n = 19.

Again, C(n,17) = C(19,17) = 19!(19−17)!.17!$\frac{19!}{\left(19-17\right)!.17!}$ = 171.

c.

C(n + 2,4) = 6 C(n,2)

Or, (n+2)!(n−2)!.4!$\frac{\left(\mathrm{n}+2\right)!}{\left(\mathrm{n}-2\right)!.4!}$ = 6. n!(n−2)!.2!$\frac{\mathrm{n}!}{\left(\mathrm{n}-2\right)!.2!}$à(n+2).(n+1).n!(n−2)!.4!$\frac{\left(\mathrm{n}+2\right).\left(\mathrm{n}+1\right).\mathrm{n}!}{\left(\mathrm{n}-2\right)!.4!}$ = 6n!(n−2)!.2$\frac{6\mathrm{n}!}{\left(\mathrm{n}-2\right)!.2}$.

Or, (n+2)(n+1)4∗3∗2∗1$\frac{\left(\mathrm{n}+2\right)\left(\mathrm{n}+1\right)}{4\ast 3\ast 2\ast 1}$ = 3

Or, n2 + 3n + 2 = 72

Or, n2 + 3n – 70 = 0

Or, (n + 10)(n – 7) = 0

Either, n = - 10 or, n = 7.

n = - 10 is not possible.

So, n = 7.