COLUMN - I COLUMN - IIA)The sum ∑k=0n ∑r=kn (−1)knCr⋅rCkar isp)1B)The sum 404C4−4C1⋅303C4+4C2⋅202C4−4C3⋅101C4+4C4 equals to (101)k, where k isq)2C)T2T3 in the expansion of (a+b)n and T3T4 in the expansion of (a+b)n+3 are equal, if n isr)5D)The remainder when 22009 is divided by 17 iss)4

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	COLUMN - I		COLUMN - II
A)	The sum $\sum_{k = 0}^{n} \sum_{r = k}^{n} {(- 1)^{k}}^{n} C_{r} \cdot^{r} C_{k} a^{r}$ is	p)	1
B)	The sum $^{404} C_{4} -^{4} C_{1} \cdot^{303} C_{4} +^{4} C_{2} \cdot^{202} C_{4} -^{4} C_{3} \cdot^{101} C_{4} +^{4} C_{4}$ equals to $(101)^{k}$ , where k is	q)	2
C)	$\frac{T_{2}}{T_{3}}$ in the expansion of $(a + b)^{n}$ and $\frac{T_{3}}{T_{4}}$ in the expansion of $(a + b)^{n + 3}$ are equal, if n is	r)	5
D)	The remainder when $2^{2009}$ is divided by 17 is	s)	4

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A - p, B - q, C - r, D - s

A - p, B - r, C - s, D - q

A - q, B - s, C - r, D - p

A - p, B - s, C - r, D - q

answer is A.

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Detailed Solution

$\sum_{k = 0}^{n} \sum_{r = k}^{n} (- 1)^{k}^{n} C_{r} \cdot^{r} C_{k} a^{r}; \sum_{r = k}^{n} (\sum_{k = 0}^{n}^{r} C_{k} (- 1)^{k}) a^{r}$

B) $(1 + x)^{4 + 4 4} C_{0} - (1 + x)^{303 4} C_{1} + (1 + x)^{202 4} C_{2} - (1 + x)^{101}$

$\begin{array}{l} ^{4} C_{3} +^{4} C_{4} - 1 \Rightarrow (1 + x)^{101} = y \\ ^{4} C_{0} y^{4} -^{4} C_{1} y^{3} +^{4} C_{2} y^{2} -^{4} C_{3} y +^{4} C_{4} - 1 \\ \Rightarrow= (y - 1)^{4} - 1 = {[{(1 + x)}^{101} - 1]}^{4} - 1 \\ \Rightarrow= {[^{101} C_{1} x +^{101} C_{2} x^{2} + \dots . .]}^{4} - 1 \\ \Rightarrow Coeff of x^{4} = (101)^{4} \end{array}$

C) $(a + b)^{n} \Rightarrow \frac{T_{2}}{T_{3}} = \frac{^{n} C_{1} a^{n - 1} b^{1}}{^{n} C_{2} a^{n - 2} b} =$

$\begin{array}{l} \frac{T_{3}}{T_{4}} = \frac{^{n + 3} C_{2} a^{n + 1} b^{2}}{^{n + 3} C_{3} a^{n} b^{3}} \Rightarrow \frac{2}{n - 1} = \frac{3}{n + 1} \\ \Rightarrow 2 n + 2 = 3 n - 3 \Rightarrow n = 5 \end{array}$