Question

Find the value of `sum_("r" = 1)^4 ""^((21 - "r"))"C"_4`

Answer 1

`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             ...[∵ ⁿC_r + ⁿC_r–1 = ⁿ⁺¹C_r   ∴ ⁿC_r–1 = ⁿ⁺¹C_r – ⁿC_r]

= ²⁰C₄ + ¹⁹C₄ + ¹⁹C₅ – ¹⁷C₅           ...[∵ ⁿC_r + ⁿC_r–1 = ⁿ⁺¹C_r]

= 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