
Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of these arrangements.
Do vowels never occur together?
A. $36000$
B. $4320$
C. $10080$
D. $40230$
Answer
576.3k+ views
Hint: The number of permutations of n objects, where p objects are of the same kind and rest are all different$ = \dfrac{{n!}}{{p!}}$.Number of permutation $ = $total number of permutation – number of permutation of all vowels occurring together
Complete step by step solution:
The given world is INDEPENDENCE.
Here, $E$ repeats$4$ times, $N$ repeats $3$ times, $D$ repeats $2$ times and $E\,\,and\,\,I$ are vowels,
As they occur together, we treat them as single object $EEEI$
Now, Arranging sets of $5$ vowels as $EEEEI$,$IEEEE,EEIEE$ and so on.
In $EEEEI;$ there are $4E$ where letter E is repeating,
Now, we use the formula:\[\dfrac{{n!}}{{{P_1}!{P_2}!{P_3}!}}\].
Total number of letters $(n) = 4 + 1 = 5$
Here, \[{P_1} = 4\]
Total arrangements $ = \dfrac{{5!}}{{4!}}$
\[ = \dfrac{{5 \times 4!}}{{4!}}\]
$ = 5$
Numbers that we need to arrange $ = 7 + 1 = 8$
Here, $N = 3\,\,and\,\,D = 2$
Since letters are repeating,
Now, we use the formula $\dfrac{{n!}}{{{P_1}{P_2}{P_3}}}$
Total letters $(n) = 8$
and ${P_1} = 3$, for letter N
${P_2} = 2$, for letter D
Total arrangement $ = \dfrac{{8!}}{{3!2!}}$
$ = 56 \times 6 \times 10$
$
= 336 \times 10 \\
= 3360 \\
$
Thus, number of arrangements $ = $total arrangements of vowels \[ \times \]total arrangements of letter
$
= 5 \times 3360 \\
= 16800 \\
$
Now, in words INDEPENDENCE
There are $3 - N({P_1})$
$4 - E({P_2})$
$2 - D({P_3})$
Total letters $ = 12$
So, $n = 12$
As letters are repeating, we use the formula as given below,
$\dfrac{{n!}}{{{P_1}!{P_2}!{P_3}!}}$
Total arrangements $ = \dfrac{{12!}}{{3!4!2!}}$
$ = \dfrac{{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4!}}{{4!\,3 \times 2 \times 1 \times 2}}$
$ = 1663200$
Number of arrangements where vowels never occur together $ = $total number of arrangement – Number of arrangements when all vowels occur together
$ = 1663200 - 16800$
$ = 1646400$
Note: In these types of questions make sure no combination of alphabets is missed and the cases of repetition too are to be solved only once.
Complete step by step solution:
The given world is INDEPENDENCE.
Here, $E$ repeats$4$ times, $N$ repeats $3$ times, $D$ repeats $2$ times and $E\,\,and\,\,I$ are vowels,
As they occur together, we treat them as single object $EEEI$
Now, Arranging sets of $5$ vowels as $EEEEI$,$IEEEE,EEIEE$ and so on.
In $EEEEI;$ there are $4E$ where letter E is repeating,
Now, we use the formula:\[\dfrac{{n!}}{{{P_1}!{P_2}!{P_3}!}}\].
Total number of letters $(n) = 4 + 1 = 5$
Here, \[{P_1} = 4\]
Total arrangements $ = \dfrac{{5!}}{{4!}}$
\[ = \dfrac{{5 \times 4!}}{{4!}}\]
$ = 5$
Numbers that we need to arrange $ = 7 + 1 = 8$
Here, $N = 3\,\,and\,\,D = 2$
Since letters are repeating,
Now, we use the formula $\dfrac{{n!}}{{{P_1}{P_2}{P_3}}}$
Total letters $(n) = 8$
and ${P_1} = 3$, for letter N
${P_2} = 2$, for letter D
Total arrangement $ = \dfrac{{8!}}{{3!2!}}$
$ = 56 \times 6 \times 10$
$
= 336 \times 10 \\
= 3360 \\
$
Thus, number of arrangements $ = $total arrangements of vowels \[ \times \]total arrangements of letter
$
= 5 \times 3360 \\
= 16800 \\
$
Now, in words INDEPENDENCE
There are $3 - N({P_1})$
$4 - E({P_2})$
$2 - D({P_3})$
Total letters $ = 12$
So, $n = 12$
As letters are repeating, we use the formula as given below,
$\dfrac{{n!}}{{{P_1}!{P_2}!{P_3}!}}$
Total arrangements $ = \dfrac{{12!}}{{3!4!2!}}$
$ = \dfrac{{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4!}}{{4!\,3 \times 2 \times 1 \times 2}}$
$ = 1663200$
Number of arrangements where vowels never occur together $ = $total number of arrangement – Number of arrangements when all vowels occur together
$ = 1663200 - 16800$
$ = 1646400$
Note: In these types of questions make sure no combination of alphabets is missed and the cases of repetition too are to be solved only once.
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