# Answer on Statistics and Probability Question for Ke_091

Question #11252

In how many ways the letters of the word ‘FARIDABAD’ can be arranged so that there are no repetitions .

Expert's answer

The word FARIDABAD is made up of three 'A's, two 'D's, one 'F', one 'R',

one

'I' and one 'B'.

First consider the three 'A's.

There are a total of 9

positions that the three 'A's can go. So by the

combination formula, the

number of combinations formed is:

9C3 (pronounced as 9 choose 3). This equals

9!/(3!(9-3)!) = 9!/3!6! =

(9*8*7*6*5*4*3*2*1)/(3*2*1)(6*5*4*3*2*1) = 504/6 =

84.

Now consider the two 'D's.

After the three 'A's have been placed

(wherever they may be), there are

6 positions left that two 'D's can go. By

the combination formula, the

number of combinations formed is:

6C2 =

6!/(2!(6-2)!) = 6!/2!4! = (6*5*4*3*2*1)/(2*1)(4*3*2*1) = 30/2 = 15.

Now

consider the one 'F'.

After the three 'A's and the two 'D's have been placed,

there are only 4

positions left that the one 'F' can go. By the combination

formula,

the number of combinations formed is:

4C1 = 4!/(1!(4-1)!) =

4!/1!3! = (4*3*2*1)/(1)(3*2*1) = 4.

Now consider the one 'R'.

After the

three 'A's, the two 'D's and the one 'F' have been placed,

there is only 3

more position left that the one 'R' can go. By the

combination formula, the

number of combinations formed is:

3C1 = 3!/(1!(3-1)!) = 3!/1!2! =

(3*2*1)/(1)(2*1) = 3.

Now consider the one 'I'.

After the three 'A's, the

two 'D's and the one 'F' and the one 'R' have been

placed, there is only 2

more position left that the one 'I' can go. By

the combination formula, the

number of combinations formed is: 2.

Lastly, consider the one 'B'.

After

the three 'A's, the two 'D's and the one 'F', one 'I' and the

one 'R' have

been

placed, there is only 1 more position left that the one

'B' can

go.

Hence, by the multiplication principle, the number of ways the

word

FARIDABAD can be arranged if all the letters are used

is:

84*15*4*3*2*1 = 30240.

one

'I' and one 'B'.

First consider the three 'A's.

There are a total of 9

positions that the three 'A's can go. So by the

combination formula, the

number of combinations formed is:

9C3 (pronounced as 9 choose 3). This equals

9!/(3!(9-3)!) = 9!/3!6! =

(9*8*7*6*5*4*3*2*1)/(3*2*1)(6*5*4*3*2*1) = 504/6 =

84.

Now consider the two 'D's.

After the three 'A's have been placed

(wherever they may be), there are

6 positions left that two 'D's can go. By

the combination formula, the

number of combinations formed is:

6C2 =

6!/(2!(6-2)!) = 6!/2!4! = (6*5*4*3*2*1)/(2*1)(4*3*2*1) = 30/2 = 15.

Now

consider the one 'F'.

After the three 'A's and the two 'D's have been placed,

there are only 4

positions left that the one 'F' can go. By the combination

formula,

the number of combinations formed is:

4C1 = 4!/(1!(4-1)!) =

4!/1!3! = (4*3*2*1)/(1)(3*2*1) = 4.

Now consider the one 'R'.

After the

three 'A's, the two 'D's and the one 'F' have been placed,

there is only 3

more position left that the one 'R' can go. By the

combination formula, the

number of combinations formed is:

3C1 = 3!/(1!(3-1)!) = 3!/1!2! =

(3*2*1)/(1)(2*1) = 3.

Now consider the one 'I'.

After the three 'A's, the

two 'D's and the one 'F' and the one 'R' have been

placed, there is only 2

more position left that the one 'I' can go. By

the combination formula, the

number of combinations formed is: 2.

Lastly, consider the one 'B'.

After

the three 'A's, the two 'D's and the one 'F', one 'I' and the

one 'R' have

been

placed, there is only 1 more position left that the one

'B' can

go.

Hence, by the multiplication principle, the number of ways the

word

FARIDABAD can be arranged if all the letters are used

is:

84*15*4*3*2*1 = 30240.

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