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# Answer to Question #11252 in Statistics and Probability 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 &#039;A&#039;s, two &#039;D&#039;s, one &#039;F&#039;, one &#039;R&#039;,
one
&#039;I&#039; and one &#039;B&#039;.
First consider the three &#039;A&#039;s.
There are a total of 9
positions that the three &#039;A&#039;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 &#039;D&#039;s.
After the three &#039;A&#039;s have been placed
(wherever they may be), there are
6 positions left that two &#039;D&#039;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 &#039;F&#039;.
After the three &#039;A&#039;s and the two &#039;D&#039;s have been placed,
there are only 4
positions left that the one &#039;F&#039; 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 &#039;R&#039;.
After the
three &#039;A&#039;s, the two &#039;D&#039;s and the one &#039;F&#039; have been placed,
there is only 3
more position left that the one &#039;R&#039; 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 &#039;I&#039;.
After the three &#039;A&#039;s, the
two &#039;D&#039;s and the one &#039;F&#039; and the one &#039;R&#039; have been
placed, there is only 2
more position left that the one &#039;I&#039; can go. By
the combination formula, the
number of combinations formed is: 2.
Lastly, consider the one &#039;B&#039;.
After
the three &#039;A&#039;s, the two &#039;D&#039;s and the one &#039;F&#039;, one &#039;I&#039; and the
one &#039;R&#039; have
been
placed, there is only 1 more position left that the one
&#039;B&#039; 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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