Question

The number of ways in which four letters of the word MATHEMATICS can be arranged is given by:
A) 136
B) 192
C) 1680
D) 2454

Answer 1

We can check the number of letters in the given word. Then we can consider 3 cases. Case 1 is the number of ways of arranging such that all the 4 letters are different. Then we can find the number of ways of arranging the 2 pairs of repeating letters and 1 pair of repeating letters and other 2 are different. The sum of the combinations of these will give the required number of ways.

Complete step by step solution:
We have the word MATHEMATICS.
It has 11 letters. Out of 11, there are 8 unique and 3 of them occur twice.
We need to arrange 4 letters from the word
Now we can consider 3 cases,
Case 1:
We can consider the case where all the 4 letters are unique.
It is given by the number of ways of selecting 4 letters from the unique 8 letters.
$\Rightarrow {N_1} = {}^8{P_4}$
We know that ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$ .
On substituting, we get
$\Rightarrow {N_1} = \dfrac{{8!}}{{\left( {8 - 4} \right)!}}$
On simplification, we get
$\Rightarrow {N_1} = \dfrac{{8!}}{{4!}}$
Now we expand the factorials.
$\Rightarrow {N_1} = \dfrac{{8 \times 7 \times 6 \times 5 \times 4!}}{{4!}}$
On simplification, we get
$\Rightarrow {N_1} = 8 \times 7 \times 6 \times 5$
On multiplication, we get
$\Rightarrow {N_1} = 1680$
Now consider case 2:
We can consider the case where 2 letters are same and 2 letters are unique.
There are 3 pairs of repeating letters. Out of them 2 can selected in ${}^3{C_2}$ ways. Then we can select the other two letters from any of the remaining 7 unique letters in ${}^7{C_2}$ ways. As two letters are repeating these four letters can be arranged in $\dfrac{{4!}}{{2!}}$ ways.
So, number of four-letter words that can be formed in case 2 is given by their product.
${N_2} = {}^3{C_1} \times {}^7{C_2} \times \dfrac{{4!}}{{2!}}$
On expanding the combinations, we get
$\Rightarrow {N_2} = \dfrac{{3!}}{{1!\left( {3 - 1} \right)!}} \times \dfrac{{7!}}{{2!\left( {7 - 2} \right)!}} \times \dfrac{{4!}}{{2!}}$
On simplification, we get
$\Rightarrow {N_2} = \dfrac{{3!}}{{2!}} \times \dfrac{{7!}}{{2! \times 5!}} \times \dfrac{{4!}}{{2!}}$
On expanding the factorial, we get
$\Rightarrow {N_2} = 3 \times \dfrac{{7 \times 6 \times 5!}}{{2 \times 5!}} \times 4 \times 3$
On cancelling the common terms, we get
$\Rightarrow {N_2} = 3 \times \dfrac{{7 \times 6}}{2} \times 4 \times 3$
On further simplification, we get
$\Rightarrow {N_2} = 756$
Case 3:
Now we can consider the case where the four letters are 2 pairs of repeating letters. 2 pairs can be selected in ${}^3{C_2}$ ways. As two letters are repeating twice these four letters can be arranged in $\dfrac{{4!}}{{2! \times 2!}}$ ways.
So, number of four-letter words that can be formed in case 2 is given by their product.
${N_3} = {}^3{C_2} \times \dfrac{{4!}}{{2! \times 2!}}$
On expanding the combinations, we get
$\Rightarrow {N_3} = \dfrac{{3!}}{{2!}} \times \dfrac{{4!}}{{2! \times 2!}}$
On expanding the factorial, we get
$\Rightarrow {N_3} = \dfrac{{3 \times 2}}{2} \times \dfrac{{4 \times 3 \times 2}}{{2 \times 2}}$
On cancelling the common terms, we get
$\Rightarrow {N_3} = 3 \times 3 \times 2$
On further simplification, we get
$\Rightarrow {N_3} = 18$
Now the total number of ways of arranging 4 letters is given by the sum of the number of ways in the three cases.
$\Rightarrow N = {N_1} + {N_2} + {N_3}$
On substituting, we get
$\Rightarrow N = 1680 + 756 + 18$
On adding, we get
$\Rightarrow N = 2454$
Therefore, the number of ways in which four letters of the word MATHEMATICS can be arranged is 2454.

So, the correct answer is option D.

Note:
We must take the repeating letters as pairs as one letter repeats only twice. We must never forget to include one of the repeating letters to the number of unique letters. We must consider all the three cases and take their sum, not the product to get the total number of ways. In the 1st case, we used the permutations instead of combinations as the order is important in arranging the letters. In the other 2 cases, after getting the possible selecting of the letters, we must multiply it with the suitable permutation in each case.