The value of $r$ for which
$S = {^{20} C}_{r} {^{10} C}_{0} + {^{20} C}_{r - 1} {^{10} C}_{1} + . . . . + {^{20} C}_{0} {^{10} C}_{r}$
is maximum is

7

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8

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10

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15

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Solution

The correct option is D $15$
This resembles the coefficient of $x^{r}$ in $(1 + x)^{20 + 10}$
$= (1 + x)^{30}$
Hence
The simplified value of the above expression is $^{30} C_{r}$ .
Now the value of any binomial coefficient $^{n} C_{r}$ is maximum when
$r = \frac{n}{2}$ when n is even and
$r = \frac{n \pm 1}{2}$ when n is odd.
Hence
In the above case
$r = \frac{30}{2} = 15$

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Similar questions

r (0 \leq r \leq 30)

{^{20} C}_{r} {^{10} C}_{0} + {^{20} C}_{r - 1} {^{10} C}_{1} + . . . . + {^{20} C}_{0} {^{10} C}_{r}

r

S =^{20} C_{r}^{10} C_{0} +^{20} C_{r - 1}^{10} C_{1} . . . . . +^{20} C_{0}^{10} C_{r}

^{40} C_{r} .^{60} C_{0} +^{40} C_{r - 1} .^{60} C_{1} + . . .

r = 50

^{2 n} C_{r}

r = n

^{40} {C_{r}}^{60} C_{0} +^{40} {C_{r - 1}}^{60} C_{1} + . . .

r = 50.

^{2 n} C_{r}

r = n

\frac{C_{0}}{1} - \frac{C_{1}}{4} + \frac{C_{2}}{7} - . . . + {(- 1)}^{n} \frac{C_{n}}{3 n + 1} =

C_{r}

{^{n} C}_{r}

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