The value of ∑r=010(−1)r 410−r 30Cr . 30−rC10−r is equal to 30Ck. 3k and ∑r=02030Cr 20Cr=nC20 then the value of (n + k) is

∑r=010(−1)r410−r 30Cr . 30−rC10−r⇒∑r=010(−1)4410−r30!r!(30−r)! . (30−r)!20!(10−r)!⇒∑r=010(−1)r . 410−r. 30C10 . 10Cr= 30C10 . 310Hence k = 10Also, ∑r=02030Cr 20Cr=∑r=02030Cr 20C20−r= 50C20∴n=50

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The value of $\sum_{r = 0}^{10} {(- 1)}^{r} 4^{10 - r}^{30} C_{r} .^{30 - r} C_{10 - r}$ is equal to $^{30} C_{k} . 3^{k}$ and ${\sum_{r = 0}^{20}}^{30} C_{r}^{20} C_{r} =^{n} C_{20}$ then the value of (n + k) is

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answer is 60.

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Detailed Solution

$\sum_{r = 0}^{10} {(- 1)}^{r} 4^{10 - r}^{30} C_{r} .^{30 - r} C_{10 - r} \Rightarrow \sum_{r = 0}^{10} {(- 1)}^{4} 4^{10 - r} \frac{30!}{r! (30 - r)!} . \frac{(30 - r)!}{20! (10 - r)!}$