Question

How many words can be formed by taking 4 different letters of the word MATHEMATICS?
A) 756
B) 1680
C) 2454
D) 18

Answer 1

Hint:
We will find different ways in which 4 letters can be selected from 11 letters of the word MATHEMATICS. Then, we will find the number of words that can be formed in each case by using the formulas of permutation and combination. Finally, we will add the number of words found in each case to find the required number of words.
Formulas used:
1) The number of \[r\]-lettered words that can be formed from \[n\] different letters is given by \[^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\] where \[n\] is the total number of letters and \[r\] is the number of letters in the word.
2) The number of \[r\]-lettered words that can be formed from \[n\] different letters where \[p\] objects of 1 type are repeated and \[q\] objects of another type are repeated is \[\dfrac{{^n{P_r}}}{{p!q!}}\].
3) The number of ways of choosing \[r\] objects from a set of \[n\] objects is \[^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}\].
4) The number of ways of choosing \[r\] objects from a set of \[n\] objects where \[p\] objects of 1 type are repeated and \[q\] objects of another type are repeated is \[\dfrac{{^n{C_r}}}{{p!q!}}\].

Complete step by step solution:
Mathematics contains 11 letters. We can choose 4 letters from it in the following ways:-
(i) 4 different letters
(ii) 2 different and 2 same letters
(iii) 2 same letters of 1 kind and 2 same letters of another kind.
(i) For the $1^{\text{st}}$ case we will have to choose 4 letters from the 8 letters M, A, T, H, E, I, C and S. We will substitute 8 for \[n\] and 4 for \[r\] in the first formula.
\[\begin{array}{l}^8{P_4} = \dfrac{{8!}}{{\left( {8 - 4} \right)!}}\\ \Rightarrow ^8{P_4} = \dfrac{{8!}}{{4!}}\end{array}\]
Evaluating the factorial, we get
\[\begin{array}{l}^8{P_4} = \dfrac{{8 \times 7 \times 6 \times 5 \times 4!}}{{4!}}\\ \Rightarrow {}^8{P_4} = 1680\end{array}\]
1680 4-letter words can be formed by using 4 different letters of MATHEMATICS.
(ii) For the $2^{\text{nd}}$ case we have to choose 1 letter which is repeated twice and 2 distinct letters from the word MATHEMATICS.
We know that the letters M, A and T are repeated twice in the given word. We have to choose 1 letter from these 3. We will find the number of ways of choosing by substituting 3 for \[n\] and 1 for \[r\] in the $3^{\text{rd}}$ the formula.
\[\begin{array}{l}^3{C_1} = \dfrac{{3!}}{{\left( {3 - 1} \right)!1!}}\\ \Rightarrow ^3{C_1} = \dfrac{{3!}}{{2!1!}}\\ \Rightarrow ^3{C_1} = 3\end{array}\]
We have to choose 2 letters from the remaining 7 letters in the word. We will now find the number of ways of choosing by substituting 7 for \[n\] and 2 for \[r\]in the $3^{\text{rd}}$ the formula.
\[\begin{array}{l}^7{C_2} = \dfrac{{7!}}{{\left( {7 - 2} \right)!2!}}\\ \Rightarrow ^7{C_2} = \dfrac{{7!}}{{5!2!}}\end{array}\]
Evaluating the factorial, we get
\[\begin{array}{l}^7{C_2} = \dfrac{{7 \times 6 \times 5!}}{{5!2!}}\\ \Rightarrow ^7{C_2} = 21\end{array}\]
We will now find the total number of words that can be formed. We will substitute 4 for \[n\], \[r\] and 2 for \[p\] in the $2^{\text{nd}}$ the formula:
\[3 \times 21 \times \dfrac{{^4{P_4}}}{{2!}} = 63 \times \dfrac{{4!}}{{\left( {4 - 4} \right)!2!}} = 756\]
The total number of 4-letter words when 1 letter which is repeated twice and 2 distinct letters from the word MATHEMATICS is 756.
(iii) We have to choose 2 same letters of 1 kind and 2 same letters of another kind from the word MATHEMATICS.
We know that the letters M, A and T are repeated twice in the given word. We have to choose 2 letters from these 3. We will find the number of ways of choosing by substituting 3 for \[n\] and 2 for \[r\]in the $3^{\text{rd}}$ formula.
\[\begin{array}{l}^3{C_2} = \dfrac{{3!}}{{\left( {3 - 2} \right)!2!}}\\ \Rightarrow \dfrac{{3!}}{{1!2!}}\\ \Rightarrow 3\end{array}\]
We will find the total number of words that can be formed. We will substitute 4 for \[n\], \[r\], 2 for \[p\] and 2 for \[q\] in the $2^{\text{nd}}$ the formula:
\[3 \times \dfrac{{^4{P_4}}}{{2!2!}} = 3 \times \dfrac{{4!}}{{\left( {4 - 4} \right)!2!2!}} = 18\]
The total number of 4-letter words when 2 same letters of 1 kind and 2 same letters of another kind form the word is 18.
The total number of words that can be formed by taking 4 different letters of the word MATHEMATICS is:
\[= 1680 + 756 + 18 = 2474\]

Option C is the correct option.

Note:
We can get confused about when to use the formula for permutations and when to use the formula for combination. A good way to remember is to check if the ordering of items matters in the given problem. If it does, we have to use the formula for permutations and if it doesn’t we have to use the formula for combination.