# Prove that 2C(n, 2) + n^2 = C(2n, 2)for n>=2?

Relevance
• Think that there are n boys and n girls (n≧2) ,

and you choose 2 persons out of them .

If you choose 2 boys , then C(n,2) ways exist .

If you choose 2 girls , then C(n,2) ways exist .

If you choose 1 boy and 1 girl , then C(n,1)*C(n,1) = n^2 ways exist .

So total of 2C(n,2) + n^2 ways exist .

On the other hand , there are C(2n,2) ways to choose 2 persons out of 2n persons .

Therefore , 2C(n,2) + n^2 = C(2n,2) .

• Log in to reply to the answers
• C(n, 2) = n!/(2!(n-2!) = (n²-n)/2

C(2n, 2) = ((2n)²-2n)/2 = (4n²-2n)/2 = 2×(n²-n)/2 + (2n²)/2 = 2×C(n, 2) + n²

This is true where nC2 exist which is n ≥ 2

• Log in to reply to the answers
• For any integer r ≥ 2,

C(r, 2) = r!/[2!(r - 2)!] = r(r - 1)(r - 2)!/[2!(r - 2)!] = r(r - 1)/2! = r(r - 1)/2

2C(n, 2) + n²

= 2[n(n - 1)/2] + n²

= 2n² - n

= (4n² - 2n)/2

= 2n(2n - 1)/2

= C(2n, 2)

• Log in to reply to the answers
• Note, C(n, r) = n!/(n-r)!r!, where n ≥ r,

2C(n, 2) + n^2 = C(2n, 2),

LHS = 2[n!/2(n-2)!] + n^2

LHS = n(n-1) + n^2 = n(2n-1),

RHS = (2n)!/(2n-2)!2! = 2n(2n-1)/2 = n(2n-1)

∴ LHS = RHS and the result is OK.

• Log in to reply to the answers