Answer :

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


M: 1 in number


O: 1 in number


R: 1 in number


A: 3 in number


D: 2 in number


B: 1 in number


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


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


iv. 3 alike letters and 1 distinct letter


i. There are 6 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 6 letters


N1 = 6C4


We know that ,


And also n! = (n)(n – 1)......2.1





N1 = 15


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 = 15 × 24


N2 = 360


ii. There are 2 letters which occurred more than once. So, we need to select 1 letter from these 2 and 2 distinct letters from the remaining 5 distinct letters. Let us assume no. of ways of selection be N3


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


N3 = (2C1) × (5C2)


We know that ,


And also n! = (n)(n – 1)......2.1





N3 = 2 × 10


N3 = 20


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 = 20 × 4 × 3


N4 = 240


iii. There are 2 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 another type


N5 = 2C2


We know that ,


And also n! = (n)(n – 1)......2.1





N5 = 1


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 = 6


iv. There is only 1 letter which occurred thrice, and 1 letter needs to be selected from the remaining 5 distinct letters.


Let us assume no. of ways of selection be N7


N7 = no. of ways of selecting 1 letter from 5 letters


N7 = 5C1


We know that ,


And also n! = (n)(n – 1)......2.1





N7 = 5


Now we need to arrange the chosen 1 different letters and 3 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 arrangements be N8.


N8 = N7 ×


N8 = 5 × 4


N8 = 20


Total no. of ways of words formed = N2 + N4 + N6 + N8


Total no. of ways of words formed = 360 + 240 + 6 + 20


Total no. of ways of words formed = 626


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