Question

Let $S_{1} = \sum\limits^{10}_{j = 1}j\left(j-1\right) ^{10}C_{j} , S_{2} = \sum\limits^{10}_{j = 1}j ^{10}C_{j}$ and $S_{3} =\sum\limits^{10}_{j = 1}j^{2} \,^{10}C_{j}.$ $S_3 = 55 \times $2^9$$ $S_1 = 90 \times 2^8$ and $S_2 = 10 \times 2^8$.

• A. Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1
• B. Statement-1 is true, Statement-2 is false
• C. Statement-1 is false, Statement-2 is true
• D. Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement - 1

Answer 1

The Correct Option is B

$S_{1} = \sum\limits^{10}_{j = 1}j\left(j-1\right) \frac{10!}{j\left(j-1\right)\left(j-2\right)!\left(10-j\right)} = 90 \sum\limits^{10}_{j = 2}\frac{8!}{\left(j-2\right)\left(8-\left(j-2\right)\right)!} = 90\cdot2^{8}.$ $S_{2} = \sum\limits^{10}_{j = 1}j \frac{10!}{j\left(j-1\right)!\left(9-\left(j-1\right)\right)!} = \sum\limits^{10}_{j = 1} \frac{9!}{\left(j-1\right)!\left(9-\left(j-1\right)\right)!} = 10\cdot2^{9}.$ $S_{3} = \sum\limits^{10}_{j = 1} \left[j \left(j-1\right)+j\right] \frac{10!}{j!\left(10-j\right)} = \sum\limits^{10}_{j = 1}j\left(j-1\right) \,^{10}C_{j} = \sum\limits^{10}_{j = 1} j\,^{10}C_{j} = 90.2^{8}+10.2^{9}$ $= 90 . 2^{8} + 20 . 2^{8} = 110 . 2^{8} = 55 . 2^{9}.$