Question

2nC3:nC2=44:3 then find n?​

Answer 1

Answer:

Step-by-step explanation:

BRAINLIEST BRAINLIEST BRAINLIEST

Answer 2

2nC3:nC2=44:3 then n is 6.

From permutation and combination, we know that:

mCr = (m!) / [r! ∗ (m−r)!]

From this above expression, we can write 2nC3 by replacing m as 2n and r as 3.

2nC3 = (2n!) / [3! * (2n-3)!]

Likewise, nC2 can also be represented by replacing m as n and r as 2.

nC2 = (n!) / [2! * (n-2)!]

So, 2nC3 : nC2 = 44:3 will be

 [(2n!)/3!∗(2n−3)!] / {(n!)/[2!∗(n−2)!]}   =   44/3   ...(1)

Also we know that the expansion of n! is:

n!   =  n ∗ (n−1) ∗ (n−2)!

So, expanding 2n! and n! will be:

2n! = 2n ∗ (2n−1) ∗ (2n−2) ∗ (2n−3)!    and     ....(2)

n! = n ∗ (n−1) ∗ (n−2)!                                         ....(3)

Replacing the values of n! and 2n! from (2) and (3) in (1), we get :

{[2n∗(2n−1)∗(2n−2)]/6} / [n∗(n−1)/2]   =    44/3            [(2n-3)! and (n-2)! in the numerator and denominator from both LHS and RHS will get divided and the value of 3! is 3*2 = 6]

Simplifying further, we get,

[2∗(2n−1)∗(2n−2)] / [3∗(n−1)] = 44/3

⇒ [(2n−1)∗(2n−2)] / [(n−1)] = 22

⇒ [ 2∗(2n−1)] = 22

⇒ 2n−1 = 11                   [Dividing both sides by 2]

⇒ 2n = 12

⇒ n = 12/2 = 6