# A five-letter word is to be formed such that the letters appearing in the odd position are taken from unrepeated letters of the word MATHEMATICS whereas the letters which occupy even places are taken amongst the repeated letters.

Last updated date: 25th Mar 2023

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Hint: Find the total number of ways to occupy even places and the total number of ways to occupy the odd places. The net total number of ways would be the product of these two results by using the fundamental theorem of arithmetic.

Unrepeated letters of the word MATHEMATICS are E, I, C, H and S. Repeated letters and their frequency of occurrence are M (2), A (2) and T (2).

In a five-letter word we have 3 odd places viz 1st, 3rd, and 5th and two even places viz 2nd and 4th.

Filling the odd places:

The number of ways in which we can fill the odd places is equivalent to the number of ways in which we can form a 3 Letter word using the letters E, I, C, H and S.

The number of ways of forming a 3 Letter word using the letters E, I, C, H and S = \[{}^{5}{{P}_{3}}\] = $\dfrac{5!}{(5-3)!}$ = $\dfrac{5!}{2!}$ = $\dfrac{5\times 4\times 3\times 2\times 1}{2\times 1}$ = $\dfrac{120}{2}$ = 60 ways

Hence the number of ways in which we can fill the odd places = 60

Filling the even places:

We can fill even places in two general ways:

[i] Both the letters are different.

[ii] Both the letters are the same.

Case I: The number of ways in which even places can be filled so that both the letters are the same is equivalent to the number of two-letter words formed using the letters M, A and T.

The number of two letter word formed using the letters M, A and T = ${}^{3}{{P}_{2}}$ = $\dfrac{3!}{(3-2)!}$ = $\dfrac{3!}{1!}$ = $\dfrac{3\times 2\times 1}{1}$ = $\dfrac{6}{1}$ = 6.

Hence the total number of ways of filling the even places so that both the letters are different = 6

Case II: We can fill the even places with both “M” s or both “A” s or both “T” s i.e. 3 possible ways

Hence the total number of ways in which even places can be filled = 6+3 = 9

Hence the total number of five-letter words such that the letters appearing at even places are the repeated letters of the word MATHEMATICS and the letters appearing at odd places are the unrepeated letters of the word MATHEMATICS = $9\times 60$= 540.

Note: In the question, we had to list out the possibilities of appearance because it involved permutation of the letters. If instead we were asked to list out the combinations, then we could have solved using the following property. The number of ways in which an “n” letter word can be formed using the collection of letters which contains “K” letters alike of one kind, “L” letters alike of other kind and rest “m” are different = coefficient of ${{x}^{n}}$ in the expansion of $\left( 1+x+{{x}^{2}}+...+{{x}^{K}} \right)\left( 1+x+{{x}^{2}}+...+{{x}^{L}} \right){{\left( 1+x \right)}^{m}}$.

__Complete step-by-step solution -__Unrepeated letters of the word MATHEMATICS are E, I, C, H and S. Repeated letters and their frequency of occurrence are M (2), A (2) and T (2).

In a five-letter word we have 3 odd places viz 1st, 3rd, and 5th and two even places viz 2nd and 4th.

Filling the odd places:

The number of ways in which we can fill the odd places is equivalent to the number of ways in which we can form a 3 Letter word using the letters E, I, C, H and S.

The number of ways of forming a 3 Letter word using the letters E, I, C, H and S = \[{}^{5}{{P}_{3}}\] = $\dfrac{5!}{(5-3)!}$ = $\dfrac{5!}{2!}$ = $\dfrac{5\times 4\times 3\times 2\times 1}{2\times 1}$ = $\dfrac{120}{2}$ = 60 ways

Hence the number of ways in which we can fill the odd places = 60

Filling the even places:

We can fill even places in two general ways:

[i] Both the letters are different.

[ii] Both the letters are the same.

Case I: The number of ways in which even places can be filled so that both the letters are the same is equivalent to the number of two-letter words formed using the letters M, A and T.

The number of two letter word formed using the letters M, A and T = ${}^{3}{{P}_{2}}$ = $\dfrac{3!}{(3-2)!}$ = $\dfrac{3!}{1!}$ = $\dfrac{3\times 2\times 1}{1}$ = $\dfrac{6}{1}$ = 6.

Hence the total number of ways of filling the even places so that both the letters are different = 6

Case II: We can fill the even places with both “M” s or both “A” s or both “T” s i.e. 3 possible ways

Hence the total number of ways in which even places can be filled = 6+3 = 9

Hence the total number of five-letter words such that the letters appearing at even places are the repeated letters of the word MATHEMATICS and the letters appearing at odd places are the unrepeated letters of the word MATHEMATICS = $9\times 60$= 540.

Note: In the question, we had to list out the possibilities of appearance because it involved permutation of the letters. If instead we were asked to list out the combinations, then we could have solved using the following property. The number of ways in which an “n” letter word can be formed using the collection of letters which contains “K” letters alike of one kind, “L” letters alike of other kind and rest “m” are different = coefficient of ${{x}^{n}}$ in the expansion of $\left( 1+x+{{x}^{2}}+...+{{x}^{K}} \right)\left( 1+x+{{x}^{2}}+...+{{x}^{L}} \right){{\left( 1+x \right)}^{m}}$.

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