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Combination

Prove that ...

Question

Prove that
$25_{4}^{C} + \sum_{r = 0}^{4} {(29 - r)}_{3}^{C} = 30_{4}^{C}$

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Solution

$^{25} C_{4} + \sum_{r = 0}^{4}^{29 - r} C_{3} =^{30} C_{4}$
Taking L.H.S.-
$^{25} C_{4} + \sum_{r = 0}^{4}^{29 - r} C_{3}$
$=^{25} C_{4} +^{29} C_{3} +^{28} C_{3} +^{27} C_{3} +^{26} C_{3} +^{25} C_{3}$
$= (^{25} C_{4} +^{25} C_{3}) +^{26} C_{3} +^{27} C_{3} +^{28} C_{3} +^{29} C_{3}$
$=^{26} C_{4} +^{26} C_{3} +^{27} C_{3} +^{28} C_{3} +^{29} C_{3} (∵^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r})$
$=^{27} C_{4} +^{27} C_{3} +^{28} C_{3} +^{29} C_{3}$
$=^{28} C_{4} +^{28} C_{3} +^{29} C_{3}$
$=^{29} C_{4} +^{29} C_{4}$
$=^{30} C_{4}$
$=$ R.H.S.
Hence proved.

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