Answer to Question #11252 in Statistics and Probability for Ke_091

Question

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

Answer 1

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.