Q. 74.0( 4 Votes )

# 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’?

Answer :

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 N_{1}

⇒ N_{1} = no. of ways of selecting 4 letters from 6 letters

⇒ N_{1} = ^{6}C_{4}

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N_{1} = 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 N_{2}

⇒ N_{2} = (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)

⇒ N_{2} = (^{3}C_{1}) × (^{5}C_{2})

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N_{2} = 3 × 10

⇒ N_{2} = 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 N_{3}

⇒ N_{3} = no. of ways of selecting 2 alike letters of one type and 2 alike letters of another type

⇒ N_{3} = ^{3}C_{2}

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N_{3} = 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 N_{4}

⇒ N_{4} = no. of ways of selecting 1 letter from 5 letters

⇒ N_{4} = ^{5}C_{1}

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N_{4} = 5

Total no. of ways of selection = N_{1} + N_{2} + N_{3} + N_{4}

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 N_{5}.

⇒ N_{5} = N_{1} × 4!

⇒ N_{5} = 15 × 24

⇒ N_{5} = 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 N_{6}.

⇒ N_{6} = N_{2} ×

⇒ N_{6} = 30 × 4 × 3

⇒ N_{6} = 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 N_{7}.

⇒ N_{7} = N_{3} ×

⇒

⇒ N_{7} = 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 N_{8}.

⇒ N_{8} = N_{4} ×

⇒ N_{8} = 5 × 4

⇒ N_{8} = 20

Total no. of ways of arrangement = N_{5} + N_{6} + N_{7} + N_{8}

Total no. of ways of arrangement = 360 + 360 + 18 + 20

Total no. of ways of arrangement = 758

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