Find the number of ways in which: (a) a selection (b) an arrangement, of four letters, can be made from the letters of the word ‘PROPORTION’?
Given the word is PROPORTION. The letters present in it are:
P: 2 in number
R: 2 in number
O: 3 in number
T: 1 in number
I: 1 in number
N: 1 in number
a. We need to find the no. of ways of selecting 4 letters from the word proportion:
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
ii. There are 3 letters which occurred more than once. So, we need to select 1 letter from these 3 and 2 distinct letters from the remaining 5 distinct letters. Let us assume no. of ways of selection be N2
⇒ N2 = (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)
⇒ N2 = (3C1) × (5C2)
We know that 
,
And also n! = (n)(n – 1)......2.1
⇒ 
⇒ 
⇒ 
⇒ N2 = 3 × 10
⇒ N2 = 30
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 N3
⇒ N3 = no. of ways of selecting 2 alike letters of one type and 2 alike letters of another type
⇒ N3 = 3C2
We know that 
,
And also n! = (n)(n – 1)......2.1
⇒ 
⇒ 
⇒ 
⇒ N3 = 3
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 N4
⇒ N4 = no. of ways of selecting 1 letter from 5 letters
⇒ N4 = 5C1
We know that 
,
And also n! = (n)(n – 1)......2.1
⇒ 
⇒ 
⇒ 
⇒ N4 = 5
Total no. of ways of selection = N1 + N2 + N3 + N4
Total no. of ways of selection = 15 + 30 + 3 + 5
Total no. of ways of selection = 53
b. We need to find the no. of ways of arranging 4 letters from the word proportion:
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. 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 N5.
⇒ N5 = N1 × 4!
⇒ N5 = 15 × 24
⇒ N5 = 360
ii. 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 N6.
⇒ N6 = N2 × 
⇒ N6 = 30 × 4 × 3
⇒ N6 = 360
iii. 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 N7.
⇒ N7 = N3 × 
⇒ 
⇒ N7 = 18
iv. 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 arrangement be N8.
⇒ N8 = N4 × 
⇒ N8 = 5 × 4
⇒ N8 = 20
Total no. of ways of arrangement = N5 + N6 + N7 + N8
Total no. of ways of arrangement = 360 + 360 + 18 + 20
Total no. of ways of arrangement = 758