Let ${S_1} = \sum\limits_{j = 1}^{10} {j\left( {j - 1} \right)} {}^{10}{C_j},{S_2} = \sum\limits_{j = 1}^{10} {j{}^{10}{C_j}} $and\[{S_3} = \sum\limits_{j = 1}^{10} {{j^2}{}^{10}{C_j}} \].
Statement-1: ${S_3} = 55 \times {2^9}$
Statement-2: ${S_1} = 90 \times {2^8}$ 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

Hint: Try to solve each of the summations one by one. Start with ${S_1}$, and use the property ${}^n{C_r} = \dfrac{n}{r}{}^{n - 1}{C_{r - 1}} = \dfrac{{n\left( {n - 1} \right)}}{{r\left( {r - 1} \right)}}{}^{n - 2}{C_{r - 2}}$ on it to simplify it. Then use ${}^n{C_0} + {}^n{C_1} + {}^n{C_2} + .......... + {}^n{C_n} = {2^n}$ to get the final value of ${S_1}$. Similarly, solve ${S_2}$ using the above two properties. In ${S_3}$, use ${j^2} = \left( {j\left( {j - 1} \right) + j} \right)$ to split the summation into two parts.

Complete step-by-step answer:
Here we have three summations ${S_1},{S_2}$and${S_3}$ to be solved. Let’s go step by step while solving each of them one by one.
We can use the property of combination that says:${}^n{C_r} = \dfrac{n}{r}{}^{n - 1}{C_{r - 1}} = \dfrac{{n\left( {n - 1} \right)}}{{r\left( {r - 1} \right)}}{}^{n - 2}{C_{r - 2}}$, in ${S_1}$
$ \Rightarrow {S_1} = \sum\limits_{j = 1}^{10} {j\left( {j - 1} \right)} {}^{10}{C_j} = \sum\limits_{j = 1}^{10} {j\left( {j - 1} \right)} \times \dfrac{{10\left( {10 - 1} \right)}}{{\left( {j - 1} \right)}} \times {}^8{C_{j - 2}}$
So, now it can be rewritten as:
\[ \Rightarrow {S_1} = \sum\limits_{j = 1}^{10} {j\left( {j - 1} \right)} \times \dfrac{{10\left( {10 - 1} \right)}}{{j\left( {j - 1} \right)}} \times {}^8{C_{j - 2}} = 9 \times 10 \times \sum\limits_{j = 1}^{10} {{}^8{C_{j - 2}}} \]
Also, we know the property: ${}^n{C_0} + {}^n{C_1} + {}^n{C_2} + .......... + {}^n{C_n} = {2^n}$
$ \Rightarrow {S_1} = 9 \times 10 \times \sum\limits_{j = 1}^{10} {{}^8{C_{j - 2}}} = 9 \times 10 \times {2^8} = 90 \times {2^8}$
Therefore, we got ${S_1} = 90 \times {2^8}$ (1)
Now, let’s move to ${S_2}$
Again using the property${}^n{C_r} = \dfrac{n}{r}{}^{n - 1}{C_{r - 1}} = \dfrac{{n\left( {n - 1} \right)}}{{r\left( {r - 1} \right)}}{}^{n - 2}{C_{r - 2}}$, in ${S_2}$
$ \Rightarrow {S_2} = \sum\limits_{j = 1}^{10} {j{}^{10}{C_j}} = \sum\limits_{j = 1}^{10} {j \times \dfrac{{10}}{j} \times {}^9{C_{j - 1}}} = 10 \times \sum\limits_{j = 1}^{10} {{}^9{C_{j - 1}}} $
Here, we can now use: ${}^n{C_0} + {}^n{C_1} + {}^n{C_2} + .......... + {}^n{C_n} = {2^n}$
\[\]
Therefore, we get \[{S_2} = 10 \times {2^9}\] (2)
Now, we move on to ${S_3}$
We can change ${j^2} = \left( {j\left( {j - 1} \right) + j} \right)$ in${S_3}$, we get:
\[ \Rightarrow {S_3} = \sum\limits_{j = 1}^{10} {{j^2}{}^{10}{C_j}} = \sum\limits_{j = 1}^{10} {\left( {j\left( {j - 1} \right) + j} \right) \times {}^{10}{C_j}} \]
Now, we can easily separate the above expression in two summations as:
\[ \Rightarrow {S_3} = \sum\limits_{j = 1}^{10} {\left( {j\left( {j - 1} \right) + j} \right) \times {}^{10}{C_j}} = \sum\limits_{j = 1}^{10} {j\left( {j - 1} \right) \times {}^{10}{C_j}} + \sum\limits_{j = 1}^{10} {j \times {}^{10}{C_j}} \]
This can be written as, by using${S_1} = \sum\limits_{j = 1}^{10} {j\left( {j - 1} \right)} {}^{10}{C_j},{S_2} = \sum\limits_{j = 1}^{10} {j{}^{10}{C_j}} $:
\[ \Rightarrow {S_3} = \sum\limits_{j = 1}^{10} {j\left( {j - 1} \right) \times {}^{10}{C_j}} + \sum\limits_{j = 1}^{10} {j \times {}^{10}{C_j}} = {S_1} + {S_2}\]
Now, we use the relation (1) and (2) in the above equation:
\[ \Rightarrow {S_3} = {S_1} + {S_2} = \left( {90 \times {2^8}} \right) + \left( {10 \times {2^9}} \right) = \left( {45 + 10} \right) \times {2^9} = 55 \times {2^9}\]
Hence, we found ${S_1} = 90 \times {2^8}$,\[{S_2} = 10 \times {2^9}\]and \[{S_3} = 55 \times {2^9}\]
So, we can say Statement-1 is correct but Statement-2 is false

Thus, the option (B) is the correct option.

Note: Try to go step by step in the solution with each of the summation one by one. Here combination ${}^n{C_r}$ can be defined as $\dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ where $n!$ represents the factorial of $n$ , i.e. the product of natural number till $n$. An alternative approach for this problem can be to start with the summation ${S_3}$ first and transform it to form the relation \[{S_3} = {S_1} + {S_2}\]. And then solve \[{S_1}\] and \[{S_2}\]