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 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

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_{2}.

⇒ N_{2} = N_{1} × 4!

⇒ N_{2} = 15 × 24

⇒ N_{2} = 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 N_{3}

⇒ N_{3} = (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)

⇒ N_{3} = (^{2}C_{1}) × (^{5}C_{2})

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N_{3} = 2 × 10

⇒ N_{3} = 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 N_{4}.

⇒ N_{4} = N_{3} ×

⇒ N_{4} = 20 × 4 × 3

⇒ N_{4} = 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 N_{5}

⇒ N_{5} = no. of ways of selecting 2 alike letters of one type and 2 alike letters of another type

⇒ N_{5} = ^{2}C_{2}

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N_{5} = 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 N_{6}.

⇒ N_{6} = N_{5} ×

⇒

⇒ N_{6} = 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 N_{7}

⇒ N_{7} = no. of ways of selecting 1 letter from 5 letters

⇒ N_{7} = ^{5}C_{1}

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N_{7} = 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 N_{8}.

⇒ N_{8} = N_{7} ×

⇒ N_{8} = 5 × 4

⇒ N_{8} = 20

Total no. of ways of words formed = N_{2} + N_{4} + N_{6} + N_{8}

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

Total no. of ways of words formed = 626

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