Class 11th

Exercise 13.1

1. 1. $\underset{x\to 3}{lim}$x + 3 = 3 + 3 = 6.

28. In a deck of 52 cards there are 4 ace cards. The required number of ways of selecting one ace card from the four = ⁴C₁ = $\frac{4!}{1!\left(4-1\right)!}$ = $\frac{4!}{3!}$ = $\frac{4*3!}{3!}$ = 4

After selecting one ace we need to select the remaining 4 card from the remaining 48 card to have a combination of 5 cards. The required number of ways

= ⁴⁸C₄

= $\frac{48!}{4!\left(48-4\right)!}$

= 1,94,580

Therefore, the total number of ways for selecting 5 card combination out of a deck of 52 cards if there is exactly one ace in each combination

= ⁴C₁*⁴⁸C₄

= 4 * 1,94,580

= 7,78,320

27. Since we are to select 3 balls from each colour in order to select 9 balls from the collections of 6, 5 and 5 balls of red, white and blue colours respectively, we can have the combination

⁶C₃ (red) *⁵C₃ (white) *⁵C₃ (blue)

= $\frac{6!}{3!\left(6-3\right)!}$ * $\frac{5!}{3!\left(5-3\right)!}$ * $\frac{5!}{3!\left(5-3\right)!}$

= 20 * 10 * 10

= 2000

26.The number of ways of selecting a team consisting of 3 boys from 5 boys and 3 girls from 4 girls is

⁵C₃*⁴C₃

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

= $\frac{20}{2}$ * $\frac{4}{1}$

= 40

25. A chord is drawn by connecting 2 points on a circle.

As we are given with 21 points on the circle, we have the following combination to find the number of chords.

24. i. ²ⁿC₃ : ⁿC₃ = 12 : 1

=> $\frac{2n!}{3!\left(2n-3\right)!}$ ÷ $\frac{n!}{3!\left(n-3\right)!}$ = $\frac{12}{1}$

=> $\frac{2n\left(2n-1\right)\left(2n-2\right)\left(2n-3\right)!}{3!\left(2n-3\right)!}$ * $\frac{3!\left(n-3\right)!}{n\left(n-1\right)\left(n-2\right)\left(n-3\right)!}$ = 12

=> $\frac{2n\left(2n-1\right)\left(2n-2\right)}{n\left(n-1\right)\left(n-2\right)}$ = 12

=> $\frac{2\left(2n-1\right)2\left(n-1\right)}{\left(n-1\right)\left(n-2\right)}$ = 12

=> 4(2n - 1) = 12(n – 2)

=> 8n – 4 = 12n – 24

=> 24 – 4 = 12n – 8n

=> 20 = 4n

=>n = $\frac{20}{4}$

=>n = 5

ii. ²ⁿC₃ : ⁿC₃ = 11 : 1

=> $\frac{2n!}{3!\left(2n-3\right)!}$ ÷ $\frac{n!}{3!\left(n-3\right)!}$ = $\frac{11}{1}$

=> $\frac{2n\left(2n-1\right)\left(2n-2\right)\left(2n-3\right)!}{3!\left(2n-3\right)!}$ * $\frac{3!\left(n-3\right)!}{n\left(n-1\right)\left(n-2\right)\left(n-3\right)!}$ = 11

=> $\frac{2n\left(2n-1\right)2\left(n-1\right)}{n\left(n-1\right)\left(n-2\right)}$ = 11

=> 4(2n – 1) = 11(n – 2)

=> 8n – 4 = 11n – 22

=> 22 – 4 = 11n – 8n

=> 18 = 3n

=>n = $\frac{18}{3}$

=>n = 6

54. Since, the lock can be open by a combination of four digits from the given ten digits I e, from 0 to 9. The number of ways of selecting 4 digits, = 10C₄

This combination of 4 digits can again be arranged within themselves in 41 ways. So, total number of

23. ⁿC₈ = ⁿC₂

As, ⁿC_{a =} ⁿC_b

=>a = b or a = n – b

=>n = a + b

We have,

ⁿC₈ = ⁿC₂

=>n = 8 + 2

=>n = 10

Therefore,

ⁿC₂

= ⁿC₂

= $\frac{10!}{2!\left(10?2\right)!}$

= $\frac{10!}{2!8!}$

= 45

52. Total no. of person selected to represent the company n (s) = 5.

Let A: person is male.

A = {Harish, Rohan, Salim}

n (A) = 3

And B: person has one 35 yes of age.

B = {Sheetal, Salim}

n (B) = 2

And A ∩ B = {Salim}

n (A ∩ B) = 1

Probability that person is either male or over 35 years.

= P (A ∪ B) = P (A) + P (B) P (A ∩ B)

$=\frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}-\frac{n(A\cap \; B)}{n(S)}$

= $=\frac{3}{5}+\frac{2}{5}-\frac{1}{5}$

$=\frac{3+2-1}{5}=\frac{4}{5}$