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Permutation and Combinations — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsPermutation and Combinations1 Mark
MCQ
If $ (1 + \text{x})^\text{ⁿ} = \text{C}_{0} +\text{C}_1\text{x} +\text{C}_2\text{ x}^² + …+ \ ^\text{C}\text{n} \text{ x}ⁿ,$ then the value of$\ ^\text{C}0^² + \ ^\text{C}1^² + \ ^\text{C}2^² + .....+ ^\text{C}\text{n}^\text{ⁿ} =\ ^{2\text{n}}\text{C}_\text{n}$ is:
  • A
    $\frac{(2\text{n})!}{(\text{n}!)}$
  • ✓
    $\frac{(2\text{n})!}{(\text{n}!\times\text{n}!)}$
  • C
    $\frac{(2\text{n})!}{(\text{n}!\times\text{n}!)2}$
  • D
    $\text{None of these}$
✓

Answer

Correct option: B.
$\frac{(2\text{n})!}{(\text{n}!\times\text{n}!)}$
Given,$ (1 + \text{x})^\text{ⁿ} = \text{C}_{0} +\text{C}_1\text{x} +\text{C}_2\text{ x}^² + …+ \ ^\text{C}\text{n} \text{ x}ⁿ .....1$
and $(1 + \text{x})^\text{n} = \ ^\text{C}0 \text{ x}^\text{n} + \ ^\text{C}1\text{x}^\text{n-1}+ \ ^\text{C}2 \text{ x}^\text{n-2} + .... \ ^\text{C}\text{r}\text{x }^\text{n-r} + … + \ ^\text{C}\text{n-1 } \text{x} + \ ^\text{C}\text{n } ... 2$
Multiply $1$ and $2,$ we get
$ (1 + \text{x})^\text{2ⁿ} = (\text{C}_{0} +\text{C}_1\text{x} +\text{C}_2\text{ x}^² + …+ \ ^\text{C}\text{n} \text{ x}^\text{n}\times)(\ ^\text{C}0\text{ x}^\text{n}+\text{C}_1\text{ x}^\text{n-1}+\text{C}_2\text{ x}^\text{n-2}+...+\text{C}_\text{r} \text{x}^\text{n-r}+...+\text{C}_\text{n-1}\text{ x}+\text{C}_\text{n})$
Now, equating the coefficient of $xn$ on both side, we get
$\ ^\text{C}0^2 + \ ^\text{C}1^2 + \ ^\text{C}2^² + ….+ \ ^\text{C}\text{n}^\text{n} = \ ^\text{2n}\text{C}_\text{n} = \frac{(2\text{n})!}{(\text{n}! × \text{n}!)}$

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