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Question

Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of these arrangements.

Do the vowels never occur together?

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Solution

In word INDEPENDENCE

There are 3N,4E, and 2D,1I,1P and 1C

Since letters are repeating so we use the formula n!p1!p2!p3!

Total letters =12

So,n=12

Since 3N,4E, and 2D

p1=3,p2=4,p3=2

Total arrangements=12!3!4!2!=1663200

There are 5 vowels in the given word
I,N,D,E,P,E,N,D,E,N,C,E

4Es and Is

They have occur together we treat them as single object

We treat EEEEI as a single object.

So our letters become EEEEIND,PNDNC

We arrange them now
Arranging 5 vowels:
Since vowels are coming together, they can be EEEEIIEEEEEEIEE and so on.
In EEEEI there are 4E

Since letter are repeating, we use the fomula=n!p1!p2!p3!

Total letter=n=5

As 4E are there,p1=4

Total arrangements=5!4!

Arranging remaining letters
Numbers we need to arrange=7+1=8

Since letter are repeating, we use this formula=n!p1!p2!p3!

Total letters=n=8

As 3N,2D

p1=3,p2=2

Total arrangements=8!3!2!

Hence the required number of arrangement

=8!3!2!×5!4!

=8×7×6×5×4×3!3!2!×5×4!4!

=8×7×6×5×2×5=16800

Number of arrangements where vowel never occur together
=Total number of arrangementNumber of arrangements when allthe vowels occur together

=166320016800=1646400

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