UPSEE Solved Model Paper 6

We can solve this problem using combination. The number of combinations of n different things taken r at a time is denoted by $\,{}^nC_r$ or C(n,r) .Thus $\,{}^nC_r$ = $\frac{n!}{r!(n-r)!}$ = $\frac{\,{}^nP_r}{r!}$ (0 ≤ r ≤ n) = $\frac{n(n-1)(n-2)\dots(n-r+1)}{r(r-1)(r-2)\dots 3\cdot2\cdot1}$ If r > n, then $\,{}^nC_r$ = 0 $\,{}^nC_0$ = 1, $\,{}^nC_1$ = n; $\,{}^nC_2$ = $\frac{n(n-1)}{2!}$ , $\,{}^nC_n$ = 1 nCx = nCy => x = y or x + y = n . If $\,{}^nC_x$ = $\,{}^nC_y$ and x ≠ y, then x+ y = n. By using this formula we can find the value of n. We have, $C(n^2 -n, 2)$ = $C(n^2 - n, 4)$ We know that $\,{}^nC_x$ = $\,{}^nC_y$ ⇒ x + y = n $n^2$ - n = 4 + 2 $n^2$ - n - 6 = 0 $n^2$ - 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.