Q. 105.0( 2 Votes )

Find the number of combinations and permutations of 4 letters taken from the word ‘EXAMINATION.’

Answer :

Given the word is EXAMINATION. The letters present in it are:


E: 1 in number


X: 1 in number


A: 2 in number


M: 1 in number


I: 2 in number


N: 2 in number


T: 1 in number


O: I in number


a. We need to find the no. of words formed by 4 letters from the word EXAMINATION:


The possible cases are the following:


i. 4 distinct letters


ii. 2 alike letters and 2 distinct letters.


iii. 2 alike letters of one type and 2 alike letters of another type


i. There are 8 different letters from which we need to select 4 letters. Let us assume no. of ways of selection be N1


N1 = no. of ways of selecting 4 letters from 8 letters


N1 = 8C4


We know that ,


And also n! = (n)(n – 1)......2.1





N1 = 70


Now we need to arrange the chosen 4 different letters. Since every word differs from other. The arrangement is similar to that of arranging n people in n places which are n! ways to arrange. So, the total no. of arrangements that can be made is 4!.


Let us assume no. of ways of arrangement be N2.


N2 = N1 × 4!


N2 = 70 × 24


N2 = 1680


ii. There are 3 letters which occurred more than once. So, we need to select 1 letter from these 2 and 2 distinct letters from the remaining 7 distinct letters. Let us assume no. of ways of selection be N3


N3 = (no. of ways of selecting 2 alike letters from the 3 types of alike letters) × (no. of ways of selecting 2 distinct letters from remaining 5 distinct letters)


N3 = (3C1) × (7C2)


We know that ,


And also n! = (n)(n – 1)......2.1





N3 = 3 × 21


N3 = 63


Now we need to arrange the chosen 2 different letters and 2 alike letters. The arrangement is similar to that of arranging n people in n places in which r are similar which are ways to arrange. So, the total no. of arrangements that can be made are .


Let us assume no. of ways of arrangement be N4.


N4 = N3 ×


N4 = 63 × 4 × 3


N4 = 756


iii. There are 3 letters which occurred more than once from which we need to select 2. Let us assume no. of ways of selection be N5


N5 = no. of ways of selecting 2 alike letters of one type and 2 alike letters of other type


N5 = 3C2


We know that ,


And also n! = (n)(n – 1)......2.1





N5 = 3


Now we need to arrange the chosen 2 alike letters of one type and 2 alike letters of another type. The arrangement is similar to that of arranging n people in n places in which r are similar of one type and m are similar of another type which are ways too arrange. So, the total no. of arrangements that can be made are .


Let us assume no. of ways of arrangement be N6.


N6 = N5 ×



N6 = 18


Total no. of combinations = N1 + N3 + N5


Total no. of combinations = 70 + 63 + 3


Total no. of combinations = 136


Total no. of ways of permutations = N2 + N4 + N6


Total no. of ways of permutations = 1680 + 756 + 18


Total no. of ways of permutations = 2454


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