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Question
Find the value of `sum_("r" = 1)^4 ""^((21 - "r"))"C"_4`
Solution
`sum_("r" = 1)^4 ""^((21 - "r"))"C"_4 = ""^((21 - 1))"C"_4 + ""^((21 - 2))"C"_4 + ""^((21 - 3))"C"_4 + ""^((21 - 4))"C"_4`
= 20C4 + 19C4 + 18C4 + 17C4
= 20C4 + 19C4 + 18C4 + 18C5 – 17C5 ...[∵ nCr + nCr–1 = n+1Cr ∴ nCr–1 = n+1Cr – nCr]
= 20C4 + 19C4 + 19C5 – 17C5 ...[∵ nCr + nCr–1 = n+1Cr]
= 20C4 + 20C5 – 17C5
= 21C5 – 17C5
= `(21!)/(5!16!) - (17!)/(5!12!)`
= `(21 xx 20 xx 19 xx 18 xx 17 xx 16!)/(5 xx 4 xx 3 xx 2 xx 1 xx 16!) - (17 xx 16 xx 15 xx 14 xx 13 xx 12!)/(5 xx 4 xx 3 xx 2 xx 1 xx 12!)`
= 21 × 19 × 3 × 17 – 17 × 2 × 14 × 13
= 20349 – 6188
= 14161
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