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?

Answer 1

In word $INDEPENDENCE$

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

Since letters are repeating so we use the formula $\frac{n!}{p_1!p_2!p_3!}$

Total letters $=12$

So,$n=12$

Since $3N,4E,$ and $2D$

$p_1=3,p_2=4,p_3=2$

Total arrangements$=\frac{12!}{3!4!2!}=1663200$

There are $5$ vowels in the given word

$I,N,D,E,P,E,N,D,E,N,C,E$

$4E^{\prime }s$ and $I^{\prime }s$

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$=\frac{n!}{p_1!p_2!p_3!}$

Total letter$=n=5$

As $4E$ are there,$p_1=4$

Total arrangements$=\frac{5!}{4!}$

Arranging remaining letters

Numbers we need to arrange$=7+1=8$

Since letter are repeating, we use this formula$=\frac{n!}{p_1!p_2!p_3!}$

Total letters$=n=8$

As $3N,2D$

$\Rightarrow p_1=3,p_2=2$

Total arrangements$=\frac{8!}{3!2!}$

Hence the required number of arrangement

$=\frac{8!}{3!2!}\times \frac{5!}{4!}$

$=\frac{8\times 7\times 6\times 5\times 4\times 3!}{3!2!}\times \frac{5\times 4!}{4!}$

$=8\times 7\times 6\times 5\times 2\times 5=16800$

Number of arrangements where vowel never occur together

$=$Total number of arrangement$-$Number of arrangements when allthe vowels occur together

$=1663200-16800=1646400$