[JEE Advanced 1980] the value of binom 47 4 sum j 1 5 binom 52 j 3
The value of $\binom{47}4+\sum_{j=1}^{5}\binom{52-j}3$ equals
(a) $\binom{47}5$
(b) $\binom{52}5$
(c) $\binom{52}4$
(d) none of these
1 Answer
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✓ Vidaara Team
✓ Accepted
· 2d ago
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Correct answer: (c) $\binom{52}4$
Repeatedly applying $\binom nr+\binom n{r-1}=\binom{n+1}r$ (hockey-stick) collapses the sum to $\binom{52}4$.
JEE Advanced 1980 · Permutations and Combinations — verified solution by the Vidaara Team.
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