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Mathematics Question on permutations and combinations

Let S1=∑j=110j(j−1)10Cj,S2=∑j=110j10CjS_{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 S3=∑j=110j2 10Cj.S_{3} =\sum\limits^{10}_{j = 1}j^{2} \,^{10}C_{j}. S3=55×S_3 = 55 \times 2^9$$ S1=90×28S_1 = 90 \times 2^8 and S2=10×28S_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

Statement-1 is true, Statement-2 is false

Explanation

Solution

S1=∑j=110j(j−1)10!j(j−1)(j−2)!(10−j)=90∑j=2108!(j−2)(8−(j−2))!=90⋅28.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}. S2=∑j=110j10!j(j−1)!(9−(j−1))!=∑j=1109!(j−1)!(9−(j−1))!=10⋅29.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}. S3=∑j=110[j(j−1)+j]10!j!(10−j)=∑j=110j(j−1) 10Cj=∑j=110j 10Cj=90.28+10.29S_{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} =90.28+20.28=110.28=55.29.= 90 . 2^{8} + 20 . 2^{8} = 110 . 2^{8} = 55 . 2^{9}.