# Question of Permutation and combination

I have found a question from somewhere in the internet as follows:

English language has 26 alphabets, out of 4 distinct vowels and 7 distinct consonants, how many letter patterns can be made with each to contain 3 distinct consonants and 2 distinct vowels?

I can't find out the solution. Can anyone help me to find it out here?

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It has 26 letters, not alphabets. Have you tried anything? Any thoughts you have? Sketch: How many ways are there to pick three consonants out of 7? How many ways are there to pick 2 vowels out of 4? How many ways are there to arrange 5 letters? – whacka Aug 16 '14 at 2:33
There are 7c3 ways to choose the 3 constants from 7 and 4c2 ways to choose 2 vowels from 4. To fill the 5 place, there could be 5! possible arrangements. So, the result will be (7c3) * (4c2) * 5!? – Khandaker Mustakimur Rahman Aug 16 '14 at 2:42

Outline: We assume that you want to make $5$-letter patterns.

$1$. How many ways are there to choose the $3$ consonants and $2$ vowels that we will use?

$2$. For every way of choosing the letters we will use, how many ways are there to line them up in a row?

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There are 7c3 ways to choose the 3 constants from 7 and 4c2 ways to choose 2 vowels from 4. To fill the 5 place, there could be 5! possible arrangements. So, the result will be (7c3) * (4c2) * 5!? – Khandaker Mustakimur Rahman Aug 16 '14 at 2:41
Yes, that is correct. – André Nicolas Aug 16 '14 at 2:42

As per a lucid conclusion, I have posted it as follows [The solution have copied from comments]:

From @whacka

It has 26 letters, not alphabets. Have you tried anything? Any thoughts you have? Sketch: How many ways are there to pick three consonants out of 7? How many ways are there to pick 2 vowels out of 4? How many ways are there to arrange 5 letters?

From @Me

There are 7c3 ways to choose the 3 constants from 7 and 4c2 ways to choose 2 vowels from 4. To fill the 5 place, there could be 5! possible arrangements. So, the result will be (7c3) * (4c2) * 5!?

From @Andre Nicolas

Yes, that is correct.

Thanks to @whacka and @Andre Nicolas for their help.

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