We know that ^nC_r=n!/r!(n r)!Given, ^3C_2Where n=3, r=2∴^nC_r=n!/r!(n r)!0ex0ex =3!/2!(3 2)!0ex0ex =3!/2!(1)!0ex0exbut 1!=10ex0ex =3!/2 ...

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Algebra of Derivatives

Evaluate: 3C2

Question

Evaluate: $^{3} C_{2}$

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Solution

We know that
$^{n} C_{r} = \frac{n!}{r! (n - r)!}$
Given, $^{3} C_{2}$
Where $n = 3, r = 2$
$∴^{n} C_{r} = \frac{n!}{r! (n - r)!} = \frac{3!}{2! (3 - 2)!} = \frac{3!}{2! (1)!} b u t 1! = 1 = \frac{3!}{2! \times 1} = \frac{3 \times 2 \times 1}{2 \times 1 \times 1} = \frac{3}{1} = 3$
Hence, the value of $^{3} C_{2}$ is $3$ .

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