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Number of words in which all the vowels are not together of the word GANESHPURI?

The options available are

  1. $21\cdot 7!$
  2. $42\cdot 8!$
  3. $84\cdot 7!$
  4. None

I have found words with vowels always together to be equal to $7!\cdot 4!$ and subtracted total number of words from it $(10!-7!×4!)$ but it seems wrong.

Also; 2:) Number of words with any two of the letters E,H and P are never together?

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  • $\begingroup$ To clarify, do you mean "it is not the case that all vowels are together" where AEUGNSHPRI counts as a valid arrangement since the I is not a part of the block with the rest of the vowels, or do you mean "it is the case that no vowel is next to any other vowel" where it would not have been a valid string since you have the A and the E adjacent among others. $\endgroup$ – JMoravitz Mar 13 '19 at 16:57
  • $\begingroup$ @JMoravitz Given the phrasing of the second question, the first interpretation seems sound. $\endgroup$ – N. F. Taussig Mar 13 '19 at 16:58
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    $\begingroup$ For the interpretation that no vowel is next to another vowel, that would have a result of $6!\times 7\times 6\times 5\times 4 = 840\times 6!$, which doesn't match any of the options either. $\endgroup$ – JMoravitz Mar 13 '19 at 17:14
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    $\begingroup$ @BJKShah which, if you would notice $\dfrac{7!}{4!3!}\cdot 4!\cdot 6! = 604800=6!\cdot 7\cdot 6\cdot 5\cdot 4$ so your answer is the same as the one I gave last week. $\endgroup$ – JMoravitz Mar 18 '19 at 12:02
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    $\begingroup$ It is the same idea as yours, except rather than arranging the vowels in a row outside, and then inserting them into the simultaneously picked holes after having arranged, I instead inserted them into the holes one at a time. Worded again another way, $6!$ to arrange the consonants. Then $7$ choices for where the a goes, $6$ choices for where the e goes, etc... $\endgroup$ – JMoravitz Mar 18 '19 at 12:20
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To start with we will have

$$ V\ C\ V\ C\ V\ C\ V $$

since we have $4$ vowels and we have to have a consonant inbetween each vowel. For this we have to pick $3$ consonants out of $6$ which we can do in

$$ \binom{6}{3} $$

different ways and we can place them in these spots in $3!$ different ways so

$$ \binom{6}{3}\cdot 3!. $$

The four vowels we can place in $4!$ different permutations into the said spots sowe have

$$ \binom{6}{3}\cdot 3!\cdot 4! $$ different combinations so far and we have to put the rest of the $3$ consonants in any place we like. There are $8$ spots for the firts one and after we placed this one we will have $9$ spots for the second one and finally $10$ spots for the last onegiving

$$ \binom{6}{3}\cdot 3!\cdot 4!\cdot 8\cdot 9\cdot 10=4\cdot 8\cdot 9\cdot 10\cdot 6!=2880\cdot 6! $$

And none of the alternatives given is this number.


For the second question you can count the number of ways putting these letters together in any combination and subtract it from all possibilities.

So I would say that first we clump these up and handle as one entity, there are $10$ different letters where we clump the given $3$ together so we have $8$ objects to order (seven letters and a clump) which we can do in

$$ 8! $$

different ways. Now inside the clump we can arrange these letters in $3!$ different ways so we have

$$ 8!\cdot3! $$

so far.

Can you continue?


EDIT

After reading the second question I realised that I answered the first question in a wrong way, I counted the number of ways where NONE of the vowels are next to eachohther.

Your approach was correct

$$ 10!-7!\cdot 4!=7!(8\cdot 9\cdot 10-4!)=7!\cdot 696 $$


EDIT after further question in a comment

So when you are done assigning letters to the pattern

$$ V\ C\ V\ C\ V\ C\ V $$

you made sure that there are no vowels together so where you put the remaining $3$ consonants does not matter. I put an $O$ to the possible places for the first of these three

$$ O\ V\ O\ C\ O\ V\ O\ C\ O\ V\ O\ C\ O\ V\ O $$

these are $8$ different spots. When you chose to put it somewhere (for the examples sake I put it somewhere)

$$ V\ C\ V\ C\ C\ V\ C\ V $$

Now you have $9$ spots for the next one, and so on.

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  • $\begingroup$ Thanks for the answer. For the first question (if question would be like second then), can you explain that selection of $2$ consonants and having $8$ and $9$ spots part? There are $3$ consonants and only three places for them, so how have you got $8$ and $9$ spots? $\endgroup$ – BJKShah Oct 10 '19 at 17:47
  • $\begingroup$ For the second question, $8!*3!$ if all three are together, $9!*2!*\binom{3}{2}$ for two letters together. So answer would be $10!-8!*3!-9!*2!*3$? $\endgroup$ – BJKShah Oct 10 '19 at 17:53
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    $\begingroup$ True! I calculated the number of consonants wrongly, you have $3$ left. I edit the question and give answer to your comment right away $\endgroup$ – Vinyl_cape_jawa Oct 10 '19 at 18:16
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    $\begingroup$ With the second one you have to be careful since now your "clump" and the third consonant may end up next to each other giving a case what we already counted when we counted the number of arrangements where all $3$ are clumped. $\endgroup$ – Vinyl_cape_jawa Oct 10 '19 at 18:26
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    $\begingroup$ Correct, I though would rather argue that you create a "2-clump" and put the third forbidden letter aside for and permute this clump with the letters. This gives rise to $\binom{3}{2}\cdot 2!\cdot 8!$ and now you place the last letter in somewhere but since it cannot be placed next to the 2-clump you only have $7$ viable spots. It gives the same answer with some other reasoning. $\endgroup$ – Vinyl_cape_jawa Oct 11 '19 at 6:43

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