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University of North Bengal
Answer
Hint: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.
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