NCERT Solutions Class 11 Maths Chapter 6 Exercise 6.4

Q1

1. If $^nC_8 = ^nC_2$, find $^nC_2$.

Property If $^nC_x = ^nC_y$, then either $x=y$ or $x+y=n$.

Apply Here $8 \neq 2$, so $n = 8 + 2 = 10$.

Find We need to find $^nC_2 = ^{10}C_2$.

Calculate $^{10}C_2 = \frac{10 \times 9}{2 \times 1} = \frac{90}{2}$

Answer: 45

Q2

2. Determine n if:
(i) $^{2n}C_3 : ^nC_3 = 12:1$
(ii) $^{2n}C_3 : ^nC_3 = 11:1$

(i) Ratio 12:1

Expand $\frac{^{2n}C_3}{^nC_3} = \frac{12}{1} \Rightarrow \frac{\frac{(2n)(2n-1)(2n-2)}{3!}}{\frac{n(n-1)(n-2)}{3!}} = 12$

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

Cancel $\frac{4(2n-1)}{n-2} = 12 \Rightarrow 2n-1 = 3(n-2)$

Solve $2n – 1 = 3n – 6 \Rightarrow n = 5$

(ii) Ratio 11:1

Expand Similar to above: $\frac{4(2n-1)}{n-2} = 11$

Solve $8n – 4 = 11n – 22 \Rightarrow 3n = 18 \Rightarrow n = 6$

(i) n=5, (ii) n=6

Q3

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

Logic A chord is formed by joining any 2 points. Order does not matter (AB is same as BA).

Formula Total points $n=21$. Select $r=2$. Formula: $^nC_r$

Calc $^{21}C_2 = \frac{21 \times 20}{2 \times 1}$

Result $21 \times 10 = 210$

210 chords

Q4

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

Boys Select 3 from 5: $^5C_3 = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10$

Girls Select 3 from 4: $^4C_3 = \frac{4 \times 3 \times 2}{3 \times 2 \times 1} = 4$

Total Multiply results (AND condition): $10 \times 4$

40 ways

Q5

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

Red 3 from 6: $^6C_3 = \frac{6 \times 5 \times 4}{6} = 20$

White 3 from 5: $^5C_3 = \frac{5 \times 4 \times 3}{6} = 10$

Blue 3 from 5: $^5C_3 = 10$

Total $20 \times 10 \times 10$

2000 ways

Q6

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

Aces 4 Aces in deck. Select 1: $^4C_1 = 4$

Others 48 Non-Aces. Select 4 (to make total 5): $^{48}C_4$

Calc $^{48}C_4 = \frac{48 \times 47 \times 46 \times 45}{24} = 2 \times 47 \times 46 \times 45 = 194580$

Total $4 \times 194580$

778,320 combinations

Q7

7. Select cricket team of 11 from 17 players (5 bowlers). Team must have exactly 4 bowlers.

Bowlers Total 5 available. Need 4. Select $^5C_4 = 5$.

Batsmen Total 12 (17-5) available. Need 7 (11-4). Select $^{12}C_7$.

Expand $^{12}C_7 = ^{12}C_5 = \frac{12 \times 11 \times 10 \times 9 \times 8}{120} = 792$

Total $5 \times 792$

3,960 ways

Q8

8. A bag contains 5 black and 6 red balls. Ways to select 2 black and 3 red balls?

Black Select 2 from 5: $^5C_2 = 10$

Red Select 3 from 6: $^6C_3 = 20$

Total $10 \times 20$

200 ways

Q9

9. Choose programme of 5 courses from 9 available. 2 specific courses are compulsory.

Compulsory 2 courses are fixed. We need to select $5 – 2 = 3$ more courses.

Available Total available reduced from 9 to $9 – 2 = 7$.

Calc Select 3 from 7: $^7C_3 = \frac{7 \times 6 \times 5}{6}$

35 ways

Permutations & Combinations