${6_{C}}_{2} =__$

Open in App

Solution

${n_{C}}_{r} = \frac{n!}{r! (n - r)!}$
$\begin{array}{rcl} {6_{C}}_{2} & = & \frac{6!}{2! (6 - 2)!} \\ = & \frac{6!}{2! \times 4!} \\ = & \frac{6 \times 5 \times 4!}{2 \times 1 \times 4!} \\ = & \frac{6 \times 5}{2} \\ = & 15 \end{array}$
Hence, ${6_{C}}_{2} = 15$

Suggest Corrections

Similar questions

6 C_{2}

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1^{m} C_{1} & ^{m + 3} C_{1} & ^{m + 6} C_{1}^{m} C_{2} & ^{m + 3} C_{2} & ^{m + 6} C_{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = 2^{α} 3^{β} 5^{γ}

α + β + γ

Join BYJU'S Learning Program