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Question

The value of $$^{47}C_4 + \displaystyle\sum_{r=1}^{5}{^{52-r}C_3}$$=


A
52C2
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B
52C3
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C
52C4
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D
52C5
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Solution

The correct option is B $$^{52}C_4$$
We know that,
$$ \displaystyle ^{n}{C}_{r}+{}^{n}{C}_{r-1}$$

$$\displaystyle =\frac{n!}{r!(n-r)!}+\frac{n!}{(r-1)!(n-r+1)!}$$

$$\displaystyle =\frac{n!(n-r+1)}{r!(n-r)!(n-r+1)}+\frac{n!r}{(r-1)!(n-r+1)!r}$$

$$\displaystyle =\frac{n!(n-r+1)}{r!(n-r+1)!}+\frac{n!r}{(n-r+1)!r!}$$

$$\displaystyle =\frac{n!}{r!(n-r+1)!}(n-r+1+r)$$

$$\displaystyle =\frac{n!}{r!(n-r+1)!}(n+1)$$

$$\displaystyle =\frac{(n+1)!}{r!(n-r+1)!}$$

$$=\displaystyle {}^{n+1}{C}_{r}$$

$$\Rightarrow \displaystyle {}^{n}{C}_{r}+{}^{n}{C}_{r-1}={}^{n+1}{C}_{r}$$
The equation is: $$({}^{47}{C}_{4}+{}^{47}{C}_{3})+{}^{48}{C}_{3}+{}^{49}{C}_{3}+{}^{50}{C}_{3}+{}^{51}{C}_{3}$$
$$=({}^{48}{C}_{4}+{}^{48}{C}_{3})+{}^{49}{C}_{3}+{}^{50}{C}_{3}+{}^{51}{C}_{3}$$
$$=({}^{49}{C}_{4}+{}^{49}{C}_{3})+{}^{50}{C}_{3}+{}^{51}{C}_{3}$$
$$=({}^{50}{C}_{4}+{}^{50}{C}_{3})+{}^{51}{C}_{3}$$
$$=({}^{51}{C}_{4}+{}^{51}{C}_{3})$$
$$={}^{52}{C}_{4}$$
Hence, option 'C' is correct.

Mathematics

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