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Combination

Simplify: 3...

Question

Simplify: ${^{34} C}_{5} + \sum_{r = 0}^{4} {^{(38 - r)} C}_{4}$ .

A

^{38} C_{4}

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B

^{39} C_{4}

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C

^{38} C_{5}

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D

^{39} C_{5}

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Solution

The correct option is B $^{39} C_{5}$
$^{34} C_{5} + 4 \sum r = 0$ $^{38 - r} C_{4}$
$=^{34} C_{5} +^{38} C_{4} +^{37} C_{4} +^{36} C_{4} +^{35} C_{4} +^{34} C_{4}$
$= (^{34} C_{5} +^{34} C_{4}) +^{35} C_{4} +^{36} C_{4} +^{37} C_{4} +^{38} C_{4}$
$=^{35} C_{5} +^{35} C_{4} +^{36} C_{4} +^{37} C_{4} +^{38} C_{4}$ (Using $^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$ )
$=^{36} C_{5} +^{36} C_{4} +^{37} C_{4} +^{38} C_{4}$
$=^{37} C_{5} +^{37} C_{4} +^{38} C_{4}$
$=^{38} C_{5} +^{38} C_{4}$
$=^{39} C_{5}$

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