How many words can be formed by taking 4 letters at a time from the letters of the word ‘MORADABAD’?
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