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Let $\mathrm{S}_1=\sum_{\mathrm{j}=1}^{10} \mathrm{j}(\mathrm{j}-1)^{10} \mathrm{C}_{\mathrm{j}}, \mathrm{S}_2=\sum_{\mathrm{j}=1}^{10} \mathrm{j}^{10} \mathrm{C}_{\mathrm{j}}$ and $\mathrm{S}_3=\sum_{\mathrm{j}=1}^{10} \mathrm{j}^2{ }^{10} \mathrm{C}_{\mathrm{j}}$.
Statement-1: $\mathrm{S}_3=55 \times 2^9$
Statement-2: $S_1=90 \times 2^8$ and $S_2=10 \times 2^8$.
Options:
Statement-1: $\mathrm{S}_3=55 \times 2^9$
Statement-2: $S_1=90 \times 2^8$ and $S_2=10 \times 2^8$.
Solution:
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Verified Answer
The correct answer is:
Statement-1 is true, Statement-2 is false
Statement-1 is true, Statement-2 is false
$S_1=\sum_{j=1}^{10} j(j-1) \frac{10 !}{j(j-1)(j-2) !(10-j) !}=90 \sum_{j=2}^{10} \frac{8 !}{(j-2) !(8-(j-2)) !}=90 \cdot 2^8$.
$S_2=\sum_{j=1}^{10} j \frac{10 !}{j(j-1) !(9-(j-1)) !}=10 \sum_{j=1}^{10} \frac{9 !}{(j-1) !(9-(j-1)) !}=10 \cdot 2^9$.
$S_3=\sum_{j=1}^{10}[j(j-1)+j] \frac{10 !}{j !(10-j) !}=\sum_{j=1}^{10} j(j-1)^{10} C_j=\sum_{j=1}^{10} j^{10} C_j=90 \cdot 2^8+10 \cdot 2^9$
$=90 \cdot 2^8+20 \cdot 2^8=110 \cdot 2^8=55 \cdot 2^9$.
$S_2=\sum_{j=1}^{10} j \frac{10 !}{j(j-1) !(9-(j-1)) !}=10 \sum_{j=1}^{10} \frac{9 !}{(j-1) !(9-(j-1)) !}=10 \cdot 2^9$.
$S_3=\sum_{j=1}^{10}[j(j-1)+j] \frac{10 !}{j !(10-j) !}=\sum_{j=1}^{10} j(j-1)^{10} C_j=\sum_{j=1}^{10} j^{10} C_j=90 \cdot 2^8+10 \cdot 2^9$
$=90 \cdot 2^8+20 \cdot 2^8=110 \cdot 2^8=55 \cdot 2^9$.
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