Question

plz give answer
given answer = 63​

Answer 1

Answer:

63

Step-by-step explanation:

$\red{\boxed{\rm Solution}}$

Given,

r! = 1 x 2 x 3 x ....... r

$\sf {}^n C_r = \frac{n!}{r!(n - r)!}$

$\sf {}^n C_1 ,{}^n C_2 ,{}^n C_3 \: are \: in \: AP$

As $\sf { }^n C_1, { }^n C_2, { }^n C_3$ are in AP

Therefore,

$\sf {}^n C_2 - {}^n C_1 = {}^n C_3 - {}^n C_2$

Therefore,

$\sf {}^n C_1 + {}^n C_3 = 2{}^n C_2$

Using the $\sf { }^n C_r$ formula given in the question,

$\sf \dfrac{n!}{1!(n - 1)!} + \dfrac{n!}{3!(n - 3)!} = \dfrac{2n!}{2!(n - 2)!}$

We know that r! = 1 x 2 x 3 ..... r

Therefore,

1! = 1

2! = 1 x 2 = 2

3! = 1 x 2 x 3 = 6

Therefore,

$\sf \dfrac{n!}{(n-1)!} + \dfrac{n!}{6(n-3)!} = \dfrac{2n!}{2(n - 2)!}$

We also know that

r! = r(r - 1)(r - 2)(r - 3)..........1

Therefore,

$\sf \dfrac{n!}{(n-1)!} + \dfrac{n!}{6(n-3)!} = \dfrac{n!}{(n-2)!}$

$\sf \dfrac{n!}{(n-3)!} \bigg[ \dfrac{1}{(n-1)(n-2)} + \dfrac{1}{6}\bigg] = \dfrac{n!}{(n-2)(n-3)!}$

$\sf \dfrac{1}{(n-1)(n-2)} + \dfrac{1}{6} = \dfrac{1}{(n-2)}$

$\sf \dfrac{6 + (n-1)(n-2)}{6(n-1)(n-2)} = \dfrac{1}{(n-2)}$

Therefore,

$\sf 6 + n^2 - 2n - n + 2 = 6n - 6 \implies n^2 - 9n + 14 = 0$

Therefore,

$\sf n^2 - 7n - 2n + 14 = 0 \implies n(n - 7) - 2(n - 7) = 0 \implies (n - 7)(n - 2) = 0$

We know that Combination will be negative if n < r which is not possible,

Therefore,

The maximum r is 3 in this case

Therefore,

n > 3

(n - 7)(n - 2) = 0

Therefore,

n = 7 or 2

As 2 is less than 3 by which combination comes out to be negative

Therefore,

n = 7

Therefore,

$\sf {}^7C_1 + {}^7C_2 + {}^7C_3$

$\sf \dfrac{7 \times 6!}{6!} + \dfrac{7 \times 6 \times 5!}{2 \times 5!} + \dfrac{7 \times 6 \times 5 \times 4!}{6 \times 4!}$

$7 + 21 + 35 = 28 + 35 = 63$