
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
511.2k+ views
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 word STATICS is used once only. Therefore, we will divide 2! by the number of possible ways.
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 word STATICS is used once only. Therefore, we will divide 2! by the number of possible ways.
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