# Answer to Question #11146 in Statistics and Probability for Ke_091

Question #11146
In how many ways the letters of the word &lsquo;BANGALORE&rsquo; can be arranged so that there are no repetitions.
The word BANGALORE is made up of two &#039;A&#039;s, one &#039;B&#039;, one &#039;N&#039;, one &#039;G&#039;,
one &#039;L&#039;
and one &#039;R&#039;.
First consider the two &#039;A&#039;s.
There are a total of 9 positions
that the two &#039;A&#039;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 &#039;B&#039;.
After the two &#039;A&#039;s have been placed
(wherever they may be), there are
7 positions left that one &#039;B&#039; 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 &#039;N&#039;.
After the two &#039;A&#039;s and the one &#039;B&#039; have been placed,
there are only 6
positions left that the one &#039;N&#039; 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;G&#039;.
After the two &#039;A&#039;s, the one &#039;B&#039; and the one &#039;N&#039; have been
placed,
there is only 5 more position left that the one &#039;G&#039; 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 &#039;L&#039;.
After the two &#039;A&#039;s, the one &#039;B&#039;, the one &#039;N&#039; and the one &#039;G&#039;
have been
placed, there is only 4 more position left that the one &#039;L&#039; 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 &#039;R&#039;.
After the two &#039;A&#039;s, the one &#039;B&#039;, the one &#039;N&#039;, the one &#039;G&#039; and
the one
&#039;L&#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:
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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