Find n if C(2n, 3): C(n, 2)::44:3
Answers
Answer:
n=6
Step-by-step explanation:
A/q,
2n,3:n,2::44:3
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
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