Question

The number of permutation of 4 letters that can be made out of the letters of the words EXAMINATION is
$\left( A \right)$ 2454
$\left( B \right)$ 2452
$\left( C \right)$ 2450
$\left( D \right)$ 1806

Answer 1

Hint – In this question use the concept that during solution as there are repeated words and we have to make only 4 letters words so there 3 possible cases arises (i.e. when we choose 4 different words, when one word is alike and other two are distinct, When two words are alike) so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given word:
EXAMINATION
In the above word there are 8 distinct words which are given below.
E – 1, X – 1, A – 2, M – 1, I – 2, N – 2, T – 1, O – 1
Now we have to make the permutation of 4 letters that can be made out of the letters of the words EXAMINATION.
Case – 1: when we choose 4 different words.
So the permutation to choose 4 different words out of 8 words is ${}^8{P_4}$
Case – 2: when one word is alike and other two are distinct
In the above words there are 3 alike words so the combination to choose 1 alike word out of 3 is ${}^3{C_1}$, and the combination to choose two different words from the remaining (8 – 1) = 7 words is ${}^7{C_2}$
Now we have to arrange these words so the number of arrangements is $\dfrac{{4!}}{{2!}}$ (divide by 2! Is it because there are one alike words).
So the permutation to choose one word is alike and other two are distinct is
$ \Rightarrow {}^3{C_1} \times {}^7{C_2} \times \dfrac{{4!}}{{2!}}$
Case – 3: When two words are alike
In the above words there are 3 alike words so the combination to choose 2 alike word out of 3 is ${}^3{C_2}$
Now we have to arrange these words so the number of arrangements is $\dfrac{{4!}}{{2! \times 2!}}$ (divide by (2!$ \times $2!) Is because there are two alike words).
So the permutation to choose two alike words is
$ \Rightarrow {}^3{C_2} \times \dfrac{{4!}}{{2! \times 2!}}$
So the total number of ways of making the permutation of 4 letters from the letters of the word EXAMINATION is
$ \Rightarrow {}^8{P_4} + {}^3{C_1} \times {}^7{C_2} \times \dfrac{{4!}}{{2!}} + {}^3{C_2} \times \dfrac{{4!}}{{2! \times 2!}}$
Now as we know that ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}{\text{ and }}{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}{\text{ }}$so use this property in the above equation we have,
$ \Rightarrow \dfrac{{8!}}{{\left( {8 - 4} \right)!}} + \dfrac{{3!}}{{1!\left( {3 - 1} \right)!}} \times \dfrac{{7!}}{{2!\left( {7 - 2} \right)!}} \times \dfrac{{4!}}{{2!}} + \dfrac{{3!}}{{2!\left( {3 - 2} \right)!}} \times \dfrac{{4!}}{{2! \times 2!}}$
Now simplify this we have,
$ \Rightarrow \dfrac{{8!}}{{4!}} + \dfrac{{3!}}{{1!.2!}} \times \dfrac{{7!}}{{2!\left( 5 \right)!}} \times \dfrac{{4!}}{{2!}} + \dfrac{{3!}}{{2!\left( 1 \right)!}} \times \dfrac{{4!}}{{2! \times 2!}}$
$ \Rightarrow \dfrac{{8.7.6.5.4!}}{{4!}} + \dfrac{{3.2!}}{{1!.2!}} \times \dfrac{{7.6.5!}}{{2!\left( 5 \right)!}} \times \dfrac{{4.3.2!}}{{2!}} + \dfrac{{3.2!}}{{2!\left( 1 \right)!}} \times \dfrac{{4.3.2!}}{{2! \times 2!}}$
$ \Rightarrow 1680 + 3 \times 21 \times 12 + 3 \times 6$
$ \Rightarrow 1680 + 756 + 18 = 2454$
So this is the required answer.
Hence option (A) is the correct answer.

Note – Whenever we face such types of questions the key concept we have to remember if there are n digits in the system in which r digits are same then the number of ways to arrange them is $\dfrac{{n!}}{{r!}}$, if there are two types of digits repeated (i.e. one type of r digits are same and another type of p digits are same), then the number of ways to arrange them is $\dfrac{{n!}}{{r!.p!}}$.