Answer

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**Hint:**In this particular question use the concept that the number of ways to arrange r objects out of n objects is given as ${}^n{P_r}$, so first fill the odd places from the letters which appear without repetition in the word MATHEMATICS using above concept, then fill the even places from the letters which appear with repetition in the same word MATHEMATICS using above concept so use these concepts to reach the solution of the question.

**Complete step by step answer:**

MATHEMATICS

In the given word letters which appear without repetition are H, E, I, C, and S

So there are 5 letters in the given word without repetition.

And the letters in the given word which appear with repetition are M, A, and T.

So there are 3 letters in the given word with repetition.

Now we have to make a 5 letter word such that the letters appearing in the odd numbered positions are taken from the letters which appear without repetition in the word MATHEMATICS. Further the letters appearing in the even-numbered positions are taken from the letters which appear with repetition in the same word MATHEMATICS.

So in a five letter word the number of odd places are ${1^{st}},{3^{rd}}{\text{ and }}{5^{th}}$.

So there are 3 odd places and the number of letters which can appear in these places are H, E, I, C, and S i.e. 5 letters.

Now as we know that the number of ways to arrange r objects out of n objects is given as ${}^n{P_r}$

So the number of ways to arrange 3 letters in odd places out of 5 is ${}^5{P_3}$.

So three odd places are filled now the remaining two even places are filed by like letters of the given word.

So there are three like letters in the given word which are M, A and T.

So the number of possible cases are given below-

Case – 1: If both of the letters are of same type

So, the number of ways to choose any one pair out of three pairs is ${}^3{C_1}$.

Case – 2: If both of the letters are of different type

As we all know that the number of ways to choose and arrange r different objects out of n different objects is given as, ${}^n{P_r}$

So, the number of ways to choose and arrange any two different letters out of three like letters is ${}^3{P_2}$

So the total number of ways to fill even places of a five letter word are the addition of the above cases

$ \Rightarrow {}^3{C_1} + {}^3{P_2}$

So the total number of ways in which the five letter word can be formed are the multiplication of above two cases i.e. the product of odd and even cases.

So the total number of ways in which the five letter word can be formed are = ${}^5{P_3} \times \left( {{}^3{C_1} + {}^3{P_2}} \right)$

Now as we know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}},{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$ so use this property in the above equation we have,

So, the number of five letter words are = $\dfrac{{5!}}{{\left( {5 - 3} \right)!}} \times \left( {\dfrac{{3!}}{{3!\left( {3 - 1} \right)!}} + \dfrac{{3!}}{{\left( {3 - 2} \right)!}}} \right)$

So, the number of five letter words are = $\dfrac{{5!}}{{2!}} \times \left( {\dfrac{{3!}}{{3!.2!}} + \dfrac{{3!}}{{1!}}} \right)$

$ \Rightarrow 60 \times \left( {3 + 6} \right) = 60\left( 9 \right) = 540$ Words.

So this is the required answer.

**So, the correct answer is “Option B”.**

**Note:**Whenever we face such types of questions the key concept we have to remember is that always recall the formula of the permutation as well as of the combination which is stated above when we choose the letters for even places from the repeated letters of the given word there are two possible cases either the even place is filled by same letter or by different letter so consider both the cases as above.

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