SEBA Class 10 Advanced Maths Chapter 6.4 Solutions: Combination

Master combinations with our comprehensive SEBA Class 10 Advanced Maths Chapter 6.4 Solutions. Here, you will get reliable SEBA Class 10 Advanced Maths Chapter 6.4 Solutions for the upcoming HSLC exam. In this article, we solve all the textbook exercises of Chapter 6.4: Combinations, from the SEBA Class 10 New Advanced Mathematics book. We provide all solutions of Chapter 6.4 in a very simple way.

See also
SEBA Class 10 Advanced Maths Chapter 6.1
SEBA Class 10 Advanced Maths Chapter 6.2
SEBA Class 10 Advanced Maths Chapter 6.3

1. Find the values:

(i) ${}^{12}C_3$

${}^{12}C_3$

$= \frac{12 \times 11 \times 10}{3 \times 2 \times 1}$

$= 2 \times 11 \times 10$

$= 220$

(ii) ${}^{14}C_{11}$

${}^{14}C_{11}$

$= {}^{14}C_{14-11}$ $\quad [\because {}^nC_r = {}^nC_{n-r}]$

$= {}^{14}C_3$

$= \frac{14 \times 13 \times 12}{3 \times 2 \times 1}$

$= 14 \times 13 \times 2$

$= 364$

(iii) ${}^{49}C_2$

${}^{49}C_2$

$= \frac{49 \times 48}{2 \times 1}$

$= 49 \times 24$

$= 1176$

(iv) ${}^{17}C_4$

${}^{17}C_4$

$= \frac{17 \times 16 \times 15 \times 14}{4 \times 3 \times 2 \times 1}$

$= 17 \times 2 \times 5 \times 14$

$= 17 \times 140$

$= 2380$

(v) ${}^{4}C_2 \times {}^{7}C_2$

${}^{4}C_2 \times {}^{7}C_2$

$= \frac{4 \times 3}{2 \times 1} \times \frac{7 \times 6}{2 \times 1}$

$= 6 \times 21$

$= 126$

(vi) ${}^{25}C_3 – {}^{21}C_3$

${}^{25}C_3 – {}^{21}C_3$

$= \frac{25 \times 24 \times 23}{3 \times 2 \times 1} – \frac{21 \times 20 \times 19}{3 \times 2 \times 1}$

$= (25 \times 4 \times 23) – (7 \times 10 \times 19)$

$= 2300 – 1330$

$= 970$

(vii) ${}^{489}C_0$

${}^{489}C_0$

$= 1 \quad [\because {}^nC_0 = 1]$

(viii) $\lfloor 5 \times {}^{4}C_2 \times {}^{8}C_3$

$\lfloor 5 \times {}^{4}C_2 \times {}^{8}C_3$

$= (5 \times 4 \times 3 \times 2 \times 1) \times \frac{4 \times 3}{2 \times 1} \times \frac{8 \times 7 \times 6}{3 \times 2 \times 1}$

$= 120 \times 6 \times 56$

$= 720 \times 56$

$= 40320$

(ix) ${}^{52}C_4$

${}^{52}C_4$

$= \frac{52 \times 51 \times 50 \times 49}{4 \times 3 \times 2 \times 1}$

$= 13 \times 17 \times 25 \times 49$

$= 221 \times 1225$

$= 270725$

(x) ${}^{26}C_2$

${}^{26}C_2$

$= \frac{26 \times 25}{2 \times 1}$

$= 13 \times 25$

$= 325$

(xi) ${}^{7}C_1 \times {}^{4}C_4 + {}^{7}C_2 \times {}^{4}C_3 + {}^{7}C_3 \times {}^{4}C_2 + {}^{7}C_4 \times {}^{4}C_1$

${}^{7}C_1 \times {}^{4}C_4 + {}^{7}C_2 \times {}^{4}C_3 + {}^{7}C_3 \times {}^{4}C_2 + {}^{7}C_4 \times {}^{4}C_1$

$= \left(7 \times 1\right) + \left(\frac{7 \times 6}{2 \times 1} \times 4\right) + \left(\frac{7 \times 6 \times 5}{3 \times 2 \times 1} \times \frac{4 \times 3}{2 \times 1}\right) + \left(\frac{7 \times 6 \times 5}{3 \times 2 \times 1} \times 4\right)$

$= 7 + (21 \times 4) + (35 \times 6) + (35 \times 4)$

$= 7 + 84 + 210 + 140$

$= 441$

(xii) ${}^{7}C_4 \times {}^{4}C_3 + {}^{9}C_4 \times {}^{5}C_3$

${}^{7}C_4 \times {}^{4}C_3 + {}^{9}C_4 \times {}^{5}C_3$

$= {}^{7}C_3 \times {}^{4}C_1 + {}^{9}C_4 \times {}^{5}C_2 \quad [\because {}^nC_r = {}^nC_{n-r}]$

$= \left(\frac{7 \times 6 \times 5}{3 \times 2 \times 1} \times 4\right) + \left(\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} \times \frac{5 \times 4}{2 \times 1}\right)$

$= (35 \times 4) + (126 \times 10)$

$= 140 + 1260$

$= 1400$

(xiii) ${}^{5}C_3 + {}^{5}C_4 + {}^{6}C_3 + {}^{7}C_5$

${}^{5}C_3 + {}^{5}C_4 + {}^{6}C_3 + {}^{7}C_5$

$= ({}^{5}C_4 + {}^{5}C_3) + {}^{6}C_3 + {}^{7}C_5 \quad [\because {}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r]$

$= {}^{6}C_4 + {}^{6}C_3 + {}^{7}C_5$

$= ({}^{6}C_4 + {}^{6}C_3) + {}^{7}C_5$

$= {}^{7}C_4 + {}^{7}C_5$

$= {}^{8}C_5$

$= {}^{8}C_{8-5}$ $\quad [\because {}^nC_r = {}^nC_{n-r}]$

$= {}^{8}C_3$

$= \frac{8 \times 7 \times 6}{3 \times 2 \times 1}$

$= 8 \times 7$

$= 56$

(xiv) ${}^{1245}C_{1244} + {}^{19}C_{19} + {}^{10}C_1$

$= \frac{\lfloor 1245}{\lfloor 1244 \cdot \lfloor 1245-1244} + \frac{\lfloor 19}{\lfloor 19 \cdot \lfloor 19-19} + \frac{\lfloor 10}{\lfloor 1 \cdot \lfloor 10-1}$

$= \frac{\lfloor 1245}{\lfloor 1244 \cdot \lfloor 1} + \frac{\lfloor 19}{\lfloor 19 \cdot \lfloor 0} + \frac{\lfloor 10}{\lfloor 1 \cdot \lfloor 9}$

$= \frac{1245 \cdot \lfloor 1244}{\lfloor 1244 \cdot \lfloor 1} + 1 + \frac{10 \cdot \lfloor 9}{\lfloor 1 \cdot \lfloor 9}$

$= 1245 + 1 + 10$

$= 1256$

(xv) ${}^{8}C_1 + {}^{8}C_2 + {}^{8}C_3 + {}^{8}C_4 + {}^{8}C_5 + {}^{8}C_6 + {}^{8}C_7 + {}^{8}C_8$

$= ({}^{8}C_2 + {}^{8}C_1) + ({}^{8}C_4 + {}^{8}C_3) + ({}^{8}C_6 + {}^{8}C_5) + ({}^{8}C_8 + {}^{8}C_7)$

$= {}^{9}C_2 + {}^{9}C_4 + {}^{9}C_6 + {}^{9}C_8$

$= {}^{9}C_2 + {}^{9}C_4 + {}^{9}C_3 + {}^{9}C_1$ $\quad [\because {}^nC_r = {}^nC_{n-r}]$

$= ({}^{9}C_2 + {}^{9}C_1) + ({}^{9}C_4 + {}^{9}C_3)$

$= {}^{10}C_2 + {}^{10}C_4$

$= \frac{10 \times 9}{2 \times 1} + \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1}$

$= 45 + 210$

$= 255$

2. Find the value of $n$ if:

(i) ${}^nC_2 = 10$

Soln :

Given,

${}^nC_2 = 10$

$\implies \frac{\lfloor n}{\lfloor 2 \cdot \lfloor n-2} = 10$

$\implies \frac{n(n-1)\lfloor n-2}{(2 \times 1)\lfloor n-2} = 10$

$\implies \frac{n(n-1)}{2} = 10$

$\implies n(n-1) = 20$

$\implies n^2 – n – 20 = 0$

$\implies n^2 – 5n + 4n – 20 = 0$

$\implies n(n-5) + 4(n-5) = 0$

$\implies (n-5)(n+4) = 0$

Either,

$n – 5 = 0 \implies n = 5$

Or,

$n + 4 = 0 \implies n = -4$ (which is not acceptable as $n$ cannot be negative)

$\therefore n = 5$

(ii) ${}^nC_3 : {}^{n-1}C_4 = 7 : 3$

Soln :

Given,

${}^nC_3 : {}^{n-1}C_4 = 7 : 3$

$\implies \frac{{}^nC_3}{{}^{n-1}C_4} = \frac{7}{3}$

$\implies \frac{\frac{\lfloor n}{\lfloor 3 \cdot \lfloor n-3}}{\frac{\lfloor n-1}{\lfloor 4 \cdot \lfloor (n-1)-4}} = \frac{7}{3}$

$\implies \frac{\lfloor n}{\lfloor 3 \cdot \lfloor n-3} \times \frac{\lfloor 4 \cdot \lfloor n-5}{\lfloor n-1} = \frac{7}{3}$

$\implies \frac{n \lfloor n-1}{\lfloor 3 \cdot (n-3)(n-4)\lfloor n-5} \times \frac{4 \times \lfloor 3 \cdot \lfloor n-5}{\lfloor n-1} = \frac{7}{3}$

$\implies \frac{4n}{(n-3)(n-4)} = \frac{7}{3}$

$\implies 12n = 7(n^2 – 7n + 12)$

$\implies 12n = 7n^2 – 49n + 84$

$\implies 7n^2 – 61n + 84 = 0$

$\implies 7n^2 – 49n – 12n + 84 = 0$

$\implies 7n(n-7) – 12(n-7) = 0$

$\implies (n-7)(7n-12) = 0$

Either,

$n – 7 = 0 \implies n = 7$

Or,

$7n – 12 = 0 \implies n = \frac{12}{7}$ (which is not an integer, so not acceptable)

$\therefore n = 7$

(iii) ${}^{2n}C_3 : {}^nC_3 = 12 : 1$

Soln :

Given,

$\frac{{}^{2n}C_3}{{}^nC_3} = \frac{12}{1}$

$\implies \frac{\lfloor 2n}{\lfloor 3 \cdot \lfloor 2n-3} \times \frac{\lfloor 3 \cdot \lfloor n-3}{\lfloor n} = 12$

$\implies \frac{2n(2n-1)(2n-2)\lfloor 2n-3}{\lfloor 2n-3} \times \frac{\lfloor n-3}{n(n-1)(n-2)\lfloor n-3} = 12$

$\implies \frac{2n(2n-1) \cdot 2(n-1)}{n(n-1)(n-2)} = 12$

$\implies \frac{4(2n-1)}{n-2} = 12$

$\implies \frac{2n-1}{n-2} = 3$

$\implies 2n – 1 = 3(n – 2)$

$\implies 2n – 1 = 3n – 6$

$\implies 3n – 2n = 6 – 1$

$\implies n = 5$

$\therefore n = 5$

(iv) ${}^nC_5 = {}^nC_{11}$

Soln :

Given,

${}^nC_5 = {}^nC_{11}$

We know that if ${}^nC_r = {}^nC_s$, then either $r = s$ or $n = r + s$.

Since $5 \ne 11$,

$\therefore n = 5 + 11$

$\implies n = 16$

$\therefore n = 16$

(v) ${}^nP_4 = 30 \cdot {}^nC_5$

Soln :

Given,

${}^nP_4 = 30 \cdot {}^nC_5$

$\implies \frac{\lfloor n}{\lfloor n-4} = 30 \times \frac{\lfloor n}{\lfloor 5 \cdot \lfloor n-5}$

$\implies \frac{1}{(n-4)\lfloor n-5} = 30 \times \frac{1}{(5 \times 4 \times 3 \times 2 \times 1)\lfloor n-5}$

$\implies \frac{1}{n-4} = \frac{30}{120}$

$\implies \frac{1}{n-4} = \frac{1}{4}$

$\implies n – 4 = 4$

$\implies n = 8$

$\therefore n = 8$

3. If ${}^nC_9 = {}^nC_6$ then find the values of ${}^nC_{12}$ and ${}^{17}C_n$.

Soln : 

Given,

${}^nC_9 = {}^nC_6$

We know that if ${}^nC_r = {}^nC_s$, then $n = r + s$.

$\therefore n = 9 + 6$

$\implies n = 15$

Now, ${}^nC_{12} = {}^{15}C_{12}$ [Since, $n = 15$]

$= {}^{15}C_{15-12} \quad [\because {}^nC_r = {}^nC_{n-r}]$

$= {}^{15}C_3$

$= \frac{15 \times 14 \times 13}{3 \times 2 \times 1}$

$= 5 \times 7 \times 13$

$= 455$

Again, ${}^{17}C_n = {}^{17}C_{15}$

$= {}^{17}C_{17-15} \quad [\because {}^nC_r = {}^nC_{n-r}]$

$= {}^{17}C_2$

$= \frac{17 \times 16}{2 \times 1}$

$= 17 \times 8$

$= 136$

$\therefore {}^nC_{12} = 455$ and ${}^{17}C_n = 136$

4. If ${}^{n-1}C_r : {}^nC_r = 2 : 3$ and ${}^nC_r : {}^{n+1}C_r = 9 : 13$, then find the values of $n$ and $r$.

Soln :

Given,

${}^{n-1}C_r : {}^nC_r = 2 : 3$

$\implies \frac{\frac{\lfloor n-1}{\lfloor r \cdot \lfloor n-1-r}}{\frac{\lfloor n}{\lfloor r \cdot \lfloor n-r}} = \frac{2}{3}$

$\implies \frac{\lfloor n-1}{\lfloor r \cdot \lfloor n-1-r} \times \frac{\lfloor r \cdot (n-r)\lfloor n-1-r}{n\lfloor n-1} = \frac{2}{3}$

$\implies \frac{n-r}{n} = \frac{2}{3}$

$\implies 3(n – r) = 2n$

$\implies 3n – 3r = 2n$

$\implies n = 3r$ — (i)

Also given,

${}^nC_r : {}^{n+1}C_r = 9 : 13$

$\implies \frac{\frac{\lfloor n}{\lfloor r \cdot \lfloor n-r}}{\frac{\lfloor n+1}{\lfloor r \cdot \lfloor n+1-r}} = \frac{9}{13}$

$\implies \frac{\lfloor n}{\lfloor r \cdot \lfloor n-r} \times \frac{\lfloor r \cdot (n+1-r)\lfloor n-r}{(n+1)\lfloor n} = \frac{9}{13}$

$\implies \frac{n+1-r}{n+1} = \frac{9}{13}$

$\implies 13(n + 1 – r) = 9(n + 1)$

$\implies 13n + 13 – 13r = 9n + 9$

$\implies 4n – 13r = -4$

$\implies 4(3r) – 13r = -4$ [ from equation (i)]

$\implies 12r – 13r = -4$

$\implies -r = -4$

$\implies r = 4$

From (i), we have

$n = 3 \times 4 = 12$

$\therefore n = 12, r = 4$

5. Prove that:

(i) ${}^nC_r + {}^{n-1}C_{r-1} + {}^{n-1}C_{r-2} = {}^{n+1}C_r$

Proof :

$\text{LHS} = {}^nC_r + ({}^{n-1}C_{r-1} + {}^{n-1}C_{r-2})$ [Since, ${}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r$]

$= {}^nC_r + {}^{(n-1+1)}C_{r-1}$

$= ({}^nC_r + {}^nC_{r-1})$

$= {}^{n+1}C_r$

$= \text{RHS}$

Hence proved.

(ii) ${}^nC_{r-1} + 2 \cdot {}^nC_{r-2} + {}^nC_{r-3} = {}^{n+2}C_{r-1}$

Proof :

$\text{LHS} = {}^nC_{r-1} + 2 \cdot {}^nC_{r-2} + {}^nC_{r-3}$

$= ({}^nC_{r-1} + {}^nC_{r-2}) + ({}^nC_{r-2} + {}^nC_{r-3}) \quad \left[\text{Since, } {}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r\right]$

$= {}^{n+1}C_{r-1} + {}^{n+1}C_{r-2}$

$= {}^{(n+1)+1}C_{r-1}$

$= {}^{n+2}C_{r-1}$

$= \text{RHS}$

Hence proved.

(iii) ${}^nC_r : {}^nC_{r-1} = (n-r+1) : r, \quad r \le n$

Proof :

$\text{L.H.S.} = \frac{{}^nC_r}{{}^nC_{r-1}}$

$= \frac{\frac{\lfloor n}{\lfloor r \cdot \lfloor n-r}}{\frac{\lfloor n}{\lfloor r-1 \cdot \lfloor n-(r-1)}}$

$= \frac{\lfloor n}{\lfloor r \cdot \lfloor n-r} \times \frac{\lfloor r-1 \cdot \lfloor n-r+1}{\lfloor n}$

$= \frac{\lfloor r-1 \cdot (n-r+1)\lfloor n-r}{r\lfloor r-1 \cdot \lfloor n-r}$

$= \frac{n-r+1}{r}$

$= (n-r+1) : r$

$= \text{R.H.S.}$

Hence proved.

(iv) ${}^nC_r = \frac{n}{r} \times {}^{n-1}C_{r-1}$

Proof :

$\text{R.H.S.} = \frac{n}{r} \times {}^{n-1}C_{r-1}$

$= \frac{n}{r} \times \frac{\lfloor n-1}{\lfloor r-1 \cdot \lfloor (n-1)-(r-1)}$

$= \frac{n}{r} \times \frac{\lfloor n-1}{\lfloor r-1 \cdot \lfloor n-r}$

$= \frac{n \lfloor n-1}{r \lfloor r-1 \cdot \lfloor n-r}$

$= \frac{\lfloor n}{\lfloor r \cdot \lfloor n-r}$

$= {}^nC_r$

$= \text{L.H.S.}$

Hence proved.

(v) $(r-2) \times {}^{n-2}C_{r-2} = (n-2) \times {}^{n-3}C_{r-3}$

Proof :

$\text{L.H.S.} = (r-2) \times {}^{n-2}C_{r-2}$

$= (r-2) \times \frac{\lfloor n-2}{\lfloor r-2 \cdot \lfloor (n-2)-(r-2)}$

$= (r-2) \times \frac{\lfloor n-2}{(r-2)\lfloor r-3 \cdot \lfloor n-r}$

$= \frac{\lfloor n-2}{\lfloor r-3 \cdot \lfloor n-r}$

$= \frac{(n-2)\lfloor n-3}{\lfloor r-3 \cdot \lfloor n-r}$

Now,

$\text{R.H.S.} = (n-2) \times {}^{n-3}C_{r-3}$

$= (n-2) \times \frac{\lfloor n-3}{\lfloor r-3 \cdot \lfloor (n-3)-(r-3)}$

$= (n-2) \times \frac{\lfloor n-3}{\lfloor r-3 \cdot \lfloor n-r}$

$= \frac{(n-2)\lfloor n-3}{\lfloor r-3 \cdot \lfloor n-r}$

$\therefore \text{L.H.S.} = \text{R.H.S.}$

Hence proved.

6: Find the value of $r$ if ${}^nP_r = 3024$ and ${}^nC_r = 126$.

Soln :

Given,

${}^nP_r = 3024$

${}^nC_r = 126$

We know that:

${}^nP_r = \lfloor r \cdot {}^nC_r$

$\implies \lfloor r = \frac{{}^nP_r}{{}^nC_r}$

$\implies \lfloor r = \frac{3024}{126}$

$\implies \lfloor r = 24$

$\implies \lfloor r = \lfloor 4$

$\implies r = 4$

$\therefore$ Required value of $r = 4$.

7. From 15 students, in how many ways 4 can be selected?

Soln :

Total number of students $= 15$

Number of students to be selected $= 4$

$\therefore$ Required number of ways $= {}^{15}C_4$

$= \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1}$

$= 15 \times 7 \times 13$

$= 1365$

Q8. From 50 students how many ways a group of 3 representatives can be selected if :

(i) a particular student is always included.

(ii) a specific student is never included.

Soln : 

Total number of students $= 50$

Number of representatives to be selected $= 3$

(i) If a particular student is always included:

Then 1 place is fixed for that student, so we have to select $3 – 1 = 2$ representatives from the remaining $50 – 1 = 49$ students.

$\therefore$ Required number of ways $= {}^{49}C_2$

$= \frac{49 \times 48}{2 \times 1}$

$= 49 \times 24$

$= 1176$

(ii) Since a specific student is never included:

Hence, excluding that student, there will be $50 – 1 = 49$ students from which we will select 3 representatives.

$\therefore$ Required number of selections $= {}^{49}C_3$

$= \frac{49 \times 48 \times 47}{3 \times 2 \times 1}$

$= 49 \times 8 \times 47$

$= 18424$

Q9. How many different teams of 11 players can be formed out of 13 players if the captain can not be excluded from the team?

Solution:

Since the captain cannot be excluded from the team, he is always included in the selection.

Hence, from the remaining $13 – 1 = 12$ players we have to select $11 – 1 = 10$ players.

$\therefore$ Total number of selections $= {}^{12}C_{10}$

$= {}^{12}C_{12-10} \quad [\because {}^nC_r = {}^nC_{n-r}]$

$= {}^{12}C_2$

$= \frac{12 \times 11}{2 \times 1}$

$= 6 \times 11$

$= 66$

$\therefore$ Required number of selections $= 66$

Q10. In how many ways can a team of 5 students be selected from 10 talented students to participate in a particular project of NASA?

Solution:

From 10 talented students, 5 students can be selected in ${}^{10}C_5$ ways.

$\therefore$ Total number of selections $= {}^{10}C_5$

$= \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1}$

$= 2 \times 3 \times 2 \times 7 \times 3$

$= 252$

$\therefore$ Required selection is 252

Q11. How many ways can 11 fruits be selected from 8 apples and 5 bananas if there should be at least three bananas in each selection?

Soln :

Given:

At least 3 bananas must be included in each selection.

The possible selections are:

$\therefore$ Total number of selections:

$= ({}^8C_8 \cdot {}^5C_3) + ({}^8C_7 \cdot {}^5C_4) + ({}^8C_6 \cdot {}^5C_5)$

$= \left(1 \times \frac{5 \times 4 \times 3}{3 \times 2 \times 1}\right) + \left(\frac{8}{1} \times \frac{5}{1}\right) + \left(\frac{8 \times 7}{2 \times 1} \times 1\right)$

$= 10 + 40 + 28$

$= 78$

$\therefore$ Required selection is 78.

Q12. How many ways can a team of 11 players be made out of 15 cricket players if a particular player can never be included in the selection.

Solution:

Since a particular player can never be included in the selection, we exclude him from the group.

Hence, from the remaining $15 – 1 = 14$ players we have to select 11 players.

$\therefore$ Total number of ways $= {}^{14}C_{11}$

$= {}^{14}C_{14-11} \quad [\because {}^nC_r = {}^nC_{n-r}]$

$= {}^{14}C_3$

$= \frac{14 \times 13 \times 12}{3 \times 2 \times 1}$

$= 14 \times 13 \times 2$

$= 364$

$\therefore$ Required number of selections is 364.

Q13. In an election 3 candidates are to be selected out of 5 candidates. How many ways a voter can vote the three candidates if he has to cast his vote for at least one of the candidates.

Soln :

Total number of candidates $= 5$.

Maximum candidates that can be selected $= 3$.

Since a voter has to cast a vote for at least one candidate, the voter may vote for 1 candidate, 2 candidates, or 3 candidates.

• Number of ways to vote for 1 candidate $= {}^5C_1$

• Number of ways to vote for 2 candidates $= {}^5C_2$

• Number of ways to vote for 3 candidates $= {}^5C_3$

$\therefore$ Required number of ways $= {}^5C_1 + {}^5C_2 + {}^5C_3$

$= 5 + \frac{5 \cdot 4}{2 \cdot 1} + \frac{5 \cdot 4}{2 \cdot 1}$

$= 5 + 10 + 10$

$= 25$

$\therefore$ Required number of ways $= 25$.

Q14. A committee of 5 members is to be formed from 4 men and 7 women. How many ways the committee can be formed if :

(i) no man is taken?

(ii) at least one man and one woman are there?

(iii) at least three men are there?

Soln :

Given:

Total men $= 4$

Total women $= 7$

Total members to be selected $= 5$

(i) When no man is taken:

All 5 members are selected from 7 women.

$\therefore$ Total number of selections $= {}^{7}C_5$

$= {}^{7}C_{7-5} \quad [\because {}^nC_r = {}^nC_{n-r}]$

$= \frac{7 \times 6}{2 \times 1} $

$= 21$

(ii) When at least one man and one woman are there: The possible combinations are:

$\therefore$ Total number of selections:

$= ({}^4C_1 \times {}^7C_4) + ({}^4C_2 \times {}^7C_3) + ({}^4C_3 \times {}^7C_2) + ({}^4C_4 \times {}^7C_1)$

$= \left(4 \times \frac{7 \times 6 \times 5}{3 \times 2 \times 1}\right) + \left(\frac{4 \times 3}{2 \times 1} \times \frac{7 \times 6 \times 5}{3 \times 2 \times 1}\right) + \left(4 \times \frac{7 \times 6}{2 \times 1}\right) + (1 \times 7)$

$= (4 \times 35) + (6 \times 35) + (4 \times 21) + 7$

$= 140 + 210 + 84 + 7$

$= 441$

(iii) When at least three men are there: The committee must contain either 3 men or 4 men:

$\therefore$ Hence total number of selections:

$= ({}^4C_3 \times {}^7C_2) + ({}^4C_4 \times {}^7C_1)$

$= \left(4 \times \frac{7 \times 6}{2 \times 1}\right) + (1 \times 7)$

$= (4 \times 21) + 7$

$= 84 + 7$

$= 91$

Q15. There are 5 black and 6 red balls in a bag. How many selections can be made taking 2 black and 3 red balls?

Soln : 

Here, there are 5 black and 6 red balls in a bag.

We can select 2 black balls from 5 black balls in ${}^5C_2$ ways.

We can select 3 red balls from 6 red balls in ${}^6C_3$ ways.

$\therefore$ Total number of selections:

$= {}^5C_2 \times {}^6C_3$

$= \frac{5 \times 4}{2 \times 1} \times \frac{6 \times 5 \times 4}{3 \times 2 \times 1}$

$= 10 \times 20$

$= 200$

Q16. In a particular course, out of 9 subjects a student has to select 5 subjects. If 2 subjects are compulsory out of 9 subjects, then how many ways a student can select his subjects?

Solution:

Here, out of 9 subjects, 2 subjects are compulsory.
Hence, from the remaining $9 – 2 = 7$ subjects we have to select $5 – 2 = 3$ subjects.
$\therefore$ Total number of selections:

$= {}^{7}C_3$

$= \frac{7 \times 6 \times 5}{3 \times 2 \times 1}$

$= 35$

Q17. Joining the vertices of a decagon :

(i) How many triangles can be obtained?

(ii) How many line segments can be obtained?

Soln : 

(i) A decagon has 10 vertices ($n = 10$) and a triangle is formed by joining any 3 vertices.

$\therefore$ Total number of triangles;

$= {}^{10}C_3$

$= \frac{10 \times 9 \times 8}{3 \times 2 \times 1}$

$= 120$

(ii) A line segment is obtained by joining any 2 vertices.

$\therefore$ Total number of line segments:

$= {}^{10}C_2$

$= \frac{10 \times 9}{2 \times 1}$

$= 45$

Q18. In a question paper, there are two sections. From the total of 10 questions each section contains 5 questions. A student has to answer 6 questions. If it is not allowed to answer more than 4 questions from a section, how many ways an examinee can select 6 questions.

Soln :

Total questions $= 10$ divided into:

• Section A $= 5$ questions

• Section B $= 5$ questions

Total questions to be answered $= 6$.

Since not more than 4 questions can be answered from any section, the selection can be made in the following ways:

$\therefore$ Total number of ways:

$= ({}^5C_4 \times {}^5C_2) + ({}^5C_3 \times {}^5C_3) + ({}^5C_2 \times {}^5C_4)$

$= \left(\frac{5}{1} \times \frac{5 \times 4}{2 \times 1}\right) + \left(\frac{5 \times 4}{2 \times 1} \times \frac{5 \times 4}{2 \times 1}\right) + \left(\frac{5 \times 4}{2 \times 1} \times \frac{5}{1}\right)$

$= (5 \times 10) + (10 \times 10) + (10 \times 5)$

$= 50 + 100 + 50$

$= 200$

$\therefore$ Required number of ways is 200.

Q19. 4 vowels and 8 consonants are given. How many words (words may not have meaning) of 5 letters can be formed taking 2 different vowels and 3 different consonants?

Soln :

Given:

Total vowels $= 4$

Total consonants $= 8$

• Number of ways to select 2 vowels from 4 vowels $= {}^4C_2 \text{ ways}$
• Number of ways to select 3 consonants from 8 consonants$= {}^8C_3 \text{ ways}$
• Total selections of 5 letters (2 vowels and 3 consonants)$= {}^4C_2 \cdot {}^8C_3\text{ ways}$
• Further, these 5 letters can be arranged among themselves to form words in $\lfloor 5$ ways.

Hence, the required number of words that can be formed:

$= {}^4C_2 \times {}^8C_3 \times \lfloor 5$

$= \frac{4 \times 3}{2 \times 1} \times \frac{8 \times 7 \times 6}{3 \times 2 \times 1} \times (5 \times 4 \times 3 \times 2 \times 1)$

$= 6 \times 56 \times 120$

$= 336 \times 120$

$= 40320$

Q20. In an almirah of a library there are 7 story books and 4 poem books. How many ways a reader can select 6 books to read if he wants to take at least 3 story books in each selection?

Soln :

Given:

Total story books $= 7$

Total poem books $= 4$

Total books to be selected $= 6$

Condition: At least 3 story books must be chosen.

The possible selections are:

$\therefore$ Total number of selections:

$= ({}^7C_3 \times {}^4C_3) + ({}^7C_4 \times {}^4C_2) + ({}^7C_5 \times {}^4C_1)$

$= \left(\frac{7 \times 6 \times 5}{3 \times 2 \times 1} \times \frac{4}{1}\right) + \left(\frac{7 \times 6 \times 5}{3 \times 2 \times 1} \times \frac{4 \times 3}{2 \times 1}\right) + \left(\frac{7 \times 6}{2 \times 1} \times \frac{4}{1}\right)$

$= (35 \times 4) + (35 \times 6) + (21 \times 4)$

$= 140 + 210 + 84$

$= 434$

$\therefore$ Required number of selections is 434.