Question

The value of the expression 47C4 + ∑(52 - j)C3 for (j= 1→ 5) is equal to

Answer 1

Answer:

The value of the expression is:

270725

Step-by-step explanation:

We are asked to find the value of the expression:

$\binom{47}{4}+\sum_{1}^{5}\binom{52-j}{3}$

i.e. we need to find the value of:

$=\binom{47}{4}+\binom{52-1}{3}+\binom{52-2}{3}+\binom{52-3}{3}+\binom{52-4}{3}+\binom{52-5}{3}\\\\\\=\binom{47}{4}+\binom{51}{3}+\binom{50}{3}+\binom{49}{3}+\binom{48}{3}+\binom{47}{3}$

We know that:

$\binom{n}{m}=\dfrac{n!}{m!\times (n-m)!}$

this means that:

⇒⇒

$\binom{47}{4}=\dfrac{47!}{4!\times (47-4)!}\\\\=\dfrac{47!}{4!\times 43!}$

$\binom{47}{4}=178365$

⇒⇒

$\binom{51}{3}=\dfrac{51!}{3!\times (51-3)!}\\\\=\dfrac{51!}{3!\times 48!}\\\\=20825$

⇒⇒

$\binom{50}{3}=\dfrac{50!}{3!\times (50-3)!}\\\\=\dfrac{50!}{3!\times 47!}\\\\=19600$

⇒⇒

$\binom{49}{3}=\dfrac{49!}{3!\times (49-3)!}\\\\=\dfrac{49!}{3!\times 46!}\\\\=18424$

⇒⇒

$\binom{48}{3}=\dfrac{48!}{3!\times (48-3)!}\\\\=\dfrac{48!}{3!\times 45!}\\\\=17296$

⇒⇒

$\binom{47}{3}=\dfrac{47!}{3!\times (47-3)!}\\\\=\dfrac{47!}{3!\times 44!}\\\\=16215$

Hence, the value of the expression is:

=178365+20825+19600+18424+17296+16215

=270725