Question

If ${{\text{W}}_{1}}\text{ and }{{\text{W}}_{2}}$ are four letter and three letter words respectively formed by using letters of the word STATICS then, how many pairs of $\left( {{\text{W}}_{1}}\text{,}{{\text{W}}_{2}} \right)$ are possible, if in any pair each letter of word STATICS is used on it
\[\begin{align}
  & \text{A}.\text{ 828} \\
 & \text{B}.\text{ 126}0 \\
 & \text{C}.\text{ 396} \\
 & \text{D}.\text{ None of these} \\
\end{align}\]

Answer 1

Hint: To solve this question, we will use the fact that, number of ways of arranging r elements from n elements is given by \[{}^{n}{{P}_{r}}=\dfrac{n!}{r!}\]
Also, if there are 't' elements repeated in n then, the number of arrangement ways is \[\dfrac{n!}{r!t!}\].

Complete step-by-step solution:
Given that, the word is STATICS.
We have ${{W}_{1}}$ as a four letter word and ${{W}_{2}}$ is a three letter word. The total number of letters in the word STATICS is 7 and we have to find the pair $\left( {{\text{W}}_{1}}\text{,}{{\text{W}}_{2}} \right)$
Forming a pair $\left( {{\text{W}}_{1}}\text{,}{{\text{W}}_{2}} \right)$ is equivalent to making any arrangements to the word STATICS and taking the word formed by the first four letter as ${{W}_{1}}$ and the word formed by the remaining as ${{W}_{2}}$
Number of ways of arranging r elements from n elements is given by \[{}^{n}{{P}_{r}}\] where, \[{}^{n}{{P}_{r}}=\dfrac{n!}{r!}\]
Also, if 't' elements are repeated again in the element n and we do not want the repetitive elements then the possible number of ways to do is \[\dfrac{n!}{r!t!}\]
We have here n = 7
As number of words in STATICS is 7 and r = 2 (as we have to select from $\left( {{\text{W}}_{1}}\text{,}{{\text{W}}_{2}} \right)$ 2 set) and t = 2 (as the letter S and T are repeated).
So, we have:
Number of arrangements is \[\Rightarrow \dfrac{7!}{2!\times 2!}\]
Opening the factorial we have,
Number of arrangement \[\Rightarrow \dfrac{7\times 6\times 5\times 4\times 3\times 2\times 1}{2\times 1\times 2\times 1\times 1}\]
Cancelling the common terms, we get:
Number of arrangement \[\begin{align}
  & \Rightarrow 7\times 3\times 5\times 4\times 3 \\
 & \Rightarrow 1260 \\
\end{align}\]
Therefore, a total of 1260 pairs are possible which is option B.

Note: The possibility of error in this question can be at the point where we have to divide 2! from a number of ways. Dividing by 2! is important because we have both S and T occurring twice in STATICS, and we need the letter of the word STATICS is used once only. Therefore, we will divide 2! by the number of possible ways.