In how many ways can the letters of the word rectangle be arranged such that all the vowels are

EDIT: I was using the spelling originally given in the title, rather than in the statement of the problem; so I have edited my answer.

There are $\displaystyle\frac{11!}{3!2!}$ total ways to arrange the letters of SLUMGULLION, and

there are $\displaystyle\frac{7!}{3!}=840$ ways to arrange the consonants L,L,L,S,M,G,N. $\;\;$ There are $4!$ of these arrangements with the L's in front of the other consonants (since we have LLL_ _ _ _ with $4!$ ways to arrange S,M,G,N)

so there are $\displaystyle\frac{4!}{840}\left(\frac{11!}{3!2!}\right)=95,040$ possible ways to do this.

Alternatively, first arrange the 7 consonants L,L,L,S,M,G,N in order with the L's in front; as above, there are $4!$ ways to do this.

Next we can place 4 spaces for the vowels between these letters, so there are $\dbinom{11}{4}$ ways to do this ${\hspace.3 in}$(since there are 4 spaces and 7 dividers).

Finally, we can arrange the vowels U,U,I,O in these spaces in $\displaystyle\frac{4!}{2!}=12$ ways;

so we have $4!\dbinom{11}{4}\cdot12=95,040$ possible arrangements.

$\begingroup$

I did this question this way :-

there are 4 consonants in the words (LGBR) and there are 7 letters in the word.

$therefore$ number of in which consonants can be arranged in relative order will be $C(7,4)$ and there will be 1 arrangement of vowels for each $C(7,4)$ choices.

$C(7,4) * 1$ = $\frac{7!}{4!(7-4)!}$ = $\frac{7!}{4!3!}$ = $35$

Is my answer correct ?

Exercise :: Permutation and Combination - General Questions

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13.

In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together?

A. 10080 B. 4989600 C. 120960 D. None of these Answer: Option C Explanation: In the word 'MATHEMATICS', we treat the vowels AEAI as one letter. Thus, we have MTHMTCS (AEAI). Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different. Number of ways of arranging these letters = 8! = 10080. (2!)(2!) Now, AEAI has 4 letters in which A occurs 2 times and the rest are different. Number of ways of arranging these letters = 4! = 12. 2! Required number of words = (10080 x 12) = 120960.

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7.

How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?

Answer: Option D Explanation: Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it. The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place. The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it. Required number of numbers = (1 x 5 x 4) = 20.

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