# Find the nu

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

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