Let $C_{r}$ stand for ${^{n} C}_{r}$ and $S (n, r) = C_{0} - C_{1} + C_{2} - C_{3} + . . . . + {(- 1)}^{r} C_{r}$

S (n, r) = 28

then

n = 9, r = 2

n = 9, r = 6

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S (n, r) = - 15

, then

n = 7, r = 3

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S (n, n) = 0

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none of these

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Solution

The correct options are
C $S (n, n) = 0$
D If $S (n, r) = 28$ then $n = 9, r = 2$ or $n = 9, r = 6$
$S (n, r) =$ coefficient of $x^{r}$ in $(1 + x + x^{2} + . . .) (C_{0} - C_{1} x + C_{2} x^{2} - . . . + {(- 1)}^{n} C_{n} x^{n})$
$=$ Coefficient of $x^{r}$ in ${(1 - x)}^{- 1} {(1 - x)}^{n} = {(- 1)}^{r} (^{n - 1} C_{r})$
Now, ${(- 1)}^{r} (^{n - 1} C_{r}) = 28$
This is possible if $n = 9, r = 2$ or $n = 9, r = 6$
$S (n, n) = \sum_{n = 0}^{n} (- 1)^{n}^{n} C_{n} = 0$
Hence, option A and C are correct.

Suggest Corrections

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