Question

Find the number of combinations and permutation of 4 letters taken from the word EXAMINATION.

Answer 1

Hint: In these types of questions we need to take different cases when all the letters are the same, two same two different, two same and other two also same, all four different.

Complete step-by-step answer:
Consider the word EXAMINATION.
There are 11 letters and out of which there are 2 A’s, 2 I’s and 2 N’s.
We have to make 4 letters.
Case 1:- When all the four letters are the same.
As in the word EXAMINATION there aren't any same letters for four times.
Case 2:- When three letters are same and one is different
As in the word EXAMINATION there aren't any same letters for three times.
Case 3:- When two letters are the same and the other two are also same.
As in the word EXAMINATION, there are three letters which are repeated twice, i.e., A, I, N.
Therefore to select these, we have $ ^3{C_2} $ and to arrange we have $ \dfrac{{4!}}{{2! \cdot 2!}} $ as both the letters are being repeated twice.
Therefore, when two letters are the same and other two are also same then the number of combination and permutation will be $ ^3{C_2} \times \dfrac{{4!}}{{2! \cdot 2!}} $ .
Case 4:- When two letters are the same and the other two are different.
As in the word EXAMINATION, there are three letters which are repeated twice, i.e., A, I, N.
Therefore to select of these repeated letters, we have $ ^3{C_1} $ and to select the non-repeated letter, we have $ ^7{C_2} $ to arrange we have $ \dfrac{{4!}}{{2!}} $ as one letter is repeated.
Therefore, when two letters are same and other two different then the number of combination and permutation will be $ ^3{C_1}{ \times ^7}{C_2} \times \dfrac{{4!}}{{2!}} $ .
Case 5: When all the letters are different.
If we omit the letters which are repeated then there will be eight letters.
Now, if we select these all letters then, we have $ ^8{C_4} $ and to arrange we have $ 4! $ .
Therefore, when all the letters are different then the number of combinations and permutation will be \[^8{C_4} \times 4!\].
So, total number permutation and combination will be $ ^3{C_2} \times \dfrac{{4!}}{{2! \cdot 2!}}{ + ^3}{C_1}{ \times ^7}{C_2} \times \dfrac{{4!}}{{2!}}{ + ^8}{C_4} \times 4! = 2454 $ .
So, the correct answer is “2454”.

Note: Permutation refers to the process of organizing into some sequence or order all the members of a set. In other words, if the set is already ordered, then the permuting process is called the rearranging of its components. The mixture is a means of choosing objects from a set, so that the order of selection does not matter (unlike permutations).