Question

Find the number of arrangements of $ 4 $ letters taken from the word EXAMINATION.
(A) $ 2454 $
(B) $ 2500 $
(C) $ 2544 $
(D) None of these

Answer 1

Hint: In this problem, to find the number of arrangements we will use the concept of permutation and combination. First we will observe the total number of different letters and repeated letters in the given word. Then, we will consider the possible cases for four letter words.

Complete step-by-step answer:
In this problem, we have to find the number of arrangements of $ 4 $ letters taken from the word EXAMINATION. In the word EXAMINATION, there are $ 8 $ different letters E, X, A, M, I, N, T, O. The letters A, I and N are repeated two times and there are five other letters E, X, M, T, O. To find total number of arrangements of $ 4 $ letters, we will consider the following possible cases:
Case 1: All letters are different (for example EXMT, AMIN, …)
To fill four places with all different letters in a four letter word, we can say that the number of arrangements of $ 8 $ letters, taking $ 4 $ at a time is equal to $ {}^8{P_4} $ . Now the value of $ {}^8{P_4} $ is obtained by using the formula $ {}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}} $ . Hence, $ {}^8{P_4} = \dfrac{{8!}}{{\left( {8 - 4} \right)!}} = \dfrac{{8!}}{{4!}} = 5 \times 6 \times 7 \times 8 = 1680 $ . Hence, we can say that the number of arrangements of $ 4 $ letters (all different) is equal to $ 1680 $ .
Case 2: Two letters are the same and other two letters are different (for example AAXM, IION, NNEX, …)
Here A, I and N are two repeated letters. One letter out of these three letters can be selected in $ {}^3{C_1} $ ways. Let us assume that letter A is selected for two places then there are only two blank spaces left. These two blank spaces can be filled in $ {}^7{P_2} $ ways because if we select the letter A then we have remaining $ 7 $ letters (E, X, M, I, N, T, O). Also the letter A selected for two places can be arranged at four places in $ {}^4{C_2} $ ways. Hence, in this case the number of arrangements of $ 4 $ letters is equal to $ {}^3{C_1} \times {}^7{P_2} \times {}^4{C_2} = 3 \times 42 \times 6 = 756 $ .
Case 3: First two letters are the same and the last two letters are same (for example, AANN, NNII, AAII, …)
Here A, I and N are two repeated letters. Two letters out of these three letters can be selected in $ {}^3{C_2} $ ways. Also the two selected letters can be arranged at four places in $ {}^4{C_2} $ ways. Hence, in this case the number of arrangements of $ 4 $ letters is equal to $ {}^3{C_2} \times {}^4{C_2} = 3 \times 6 = 18 $ .
Therefore, by considering all above cases, we can say that the total number of arrangements of $ 4 $ letters is equal to $ 1680 + 756 + 18 = 2454 $ . Hence, option A is correct.
So, the correct answer is “Option A”.

Note: In this problem, we cannot consider the case of all the same letters because there are no four repeated letters in the word EXAMINATION. Similarly, we cannot consider the case of three same letters because there are no three repeated letters in the given word.