We can solve this problem using combination. The number of combinations of n different things taken r at a time is denoted by
n
‌
Cr or C(n,r) .Thus
n
‌
Cr =
n!
r!(n−r)!
=
n
‌
Pr
r!
(0 ≤ r ≤ n) =
n(n−1)(n−2)...(n−r+1)
r(r−1)(r−2)...3.2.1
If r > n, then
n
‌
Cr = 0
n
‌
C0 = 1,
n
‌
C1 = n;
n
‌
C2 =
n(n−1)
2!
,
n
‌
Cn = 1 nCx = nCy => x = y or x + y = n . If
n
‌
Cx =
n
‌
Cy and x ≠y, then x+ y = n. By using this formula we can find the value of n. We have, C(n2−n,2) = C(n2−n,4) We know that
n
‌
Cx =
n
‌
Cy ⇒ x + y = n n2 - n = 4 + 2 n2 - n - 6 = 0 n2 - 3n + 2n - 6 = 0 n(n - 3) + 2(n - 3) = 0 (n + 2) (n - 3) = 0 (n + 2) = 0 or (n - 3) = 0 n = -2 or n = 3 As 'n' cannot be negative, so n = 3. Hence, the value of n is 3.