Question #11146

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

Expert's answer

The word BANGALORE is made up of two 'A's, one 'B', one 'N', one 'G',

one 'L'

and one 'R'.

First consider the two 'A's.

There are a total of 9 positions

that the two 'A's can go. So by the

combination formula, the number of

combinations formed is:

9C2 (pronounced as 9 choose 2). This equals

9!/(2!(9-2)!) = 9!/2!7! =

(9*8*7*6*5*4*3*2*1)/(2*1)(7*6*5*4*3*2*1) = 72/2 =

36.

Now consider the one 'B'.

After the two 'A's have been placed

(wherever they may be), there are

7 positions left that one 'B' can go. By

the combination formula, the

number of combinations formed is:

7C2 =

7!/(2!(7-2)!) = 7!/2!5! = (7*6*5*4*3*2*1)/(2*1)(5*4*3*2*1) = 42/2 = 21.

Now

consider the one 'N'.

After the two 'A's and the one 'B' have been placed,

there are only 6

positions left that the one 'N' 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

'G'.

After the two 'A's, the one 'B' and the one 'N' have been

placed,

there is only 5 more position left that the one 'G' can go. By

the

combination formula, the number of combinations formed is:

5C2 =

5!/(2!(5-2)!) = 5!/2!3! = (5*4*3*2*1)/(2*1)(3*2*1) = 20/2 = 10.

Now consider

the one 'L'.

After the two 'A's, the one 'B', the one 'N' and the one 'G'

have been

placed, there is only 4 more position left that the one 'L' can go.

By

the combination formula, the number of combinations formed is:

4C2 =

4!/(2!(4-2)!) = 4!/2!2! = (4*3*2*1)/(2*1)(2*1) = 12/2 = 6.

Lastly, consider

the one 'R'.

After the two 'A's, the one 'B', the one 'N', the one 'G' and

the one

'L' 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:

3C2 = 3!/(2!(3-2)!) = 3!/2!1! = (3*2*1)/(2*1)(1) =

3.

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

word

BANGALORE can be arranged if all the letters are used

is:

36*21*15*10*6*3 = 2041200.

one 'L'

and one 'R'.

First consider the two 'A's.

There are a total of 9 positions

that the two 'A's can go. So by the

combination formula, the number of

combinations formed is:

9C2 (pronounced as 9 choose 2). This equals

9!/(2!(9-2)!) = 9!/2!7! =

(9*8*7*6*5*4*3*2*1)/(2*1)(7*6*5*4*3*2*1) = 72/2 =

36.

Now consider the one 'B'.

After the two 'A's have been placed

(wherever they may be), there are

7 positions left that one 'B' can go. By

the combination formula, the

number of combinations formed is:

7C2 =

7!/(2!(7-2)!) = 7!/2!5! = (7*6*5*4*3*2*1)/(2*1)(5*4*3*2*1) = 42/2 = 21.

Now

consider the one 'N'.

After the two 'A's and the one 'B' have been placed,

there are only 6

positions left that the one 'N' 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

'G'.

After the two 'A's, the one 'B' and the one 'N' have been

placed,

there is only 5 more position left that the one 'G' can go. By

the

combination formula, the number of combinations formed is:

5C2 =

5!/(2!(5-2)!) = 5!/2!3! = (5*4*3*2*1)/(2*1)(3*2*1) = 20/2 = 10.

Now consider

the one 'L'.

After the two 'A's, the one 'B', the one 'N' and the one 'G'

have been

placed, there is only 4 more position left that the one 'L' can go.

By

the combination formula, the number of combinations formed is:

4C2 =

4!/(2!(4-2)!) = 4!/2!2! = (4*3*2*1)/(2*1)(2*1) = 12/2 = 6.

Lastly, consider

the one 'R'.

After the two 'A's, the one 'B', the one 'N', the one 'G' and

the one

'L' 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:

3C2 = 3!/(2!(3-2)!) = 3!/2!1! = (3*2*1)/(2*1)(1) =

3.

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

word

BANGALORE can be arranged if all the letters are used

is:

36*21*15*10*6*3 = 2041200.

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