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How many words, with or without meaning, each of $2$ vowels and $3$ consonants can be formed from the letters of the word DAUGHTER?

Let me tell my combinations approach which got me the right answer:

The given word has $3$ vowels and $5$ consonants.

Number of possible combinations = $^2C_3 \times ^3C_5$. These combinations can be arranged among themselves in $5!$ ways. Hence, number of words = $5 ! \times ^2C_3 \times ^3C_5 = 3600$

Attempt using Permutations:

There are $5$ vacant places. Thus, number of permutations possible = $2!\times 3! \times^3P_2 \times ^5P_3 $. This doesn't give the right answer.

Please tell me how to approach this problem using Permutations only.

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The multiplier for your permutation values is $\binom 52$ (or equally $\binom 53$), representing the choice of positions for the vowels (or consonants).

$\binom 52 \times {}^3P_2 \times {}^5P_3$

This bring the two approaches back into agreement.

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  • $\begingroup$ Hey! Can you please tell me what that long circular bracket means? Sorry for such a silly question but I am new to combinatorics. $\endgroup$ – Abcd Dec 17 '17 at 21:04
  • $\begingroup$ $\binom 52 = {}^2C_5$ in your notation above. $\endgroup$ – Joffan Dec 17 '17 at 21:05
  • $\begingroup$ But we are talking about permutations, aren't we? $\endgroup$ – Abcd Dec 17 '17 at 21:06
  • $\begingroup$ Your permutations of vowels and of consonants stand unchanged. How to interleave them is a combinations value. $\endgroup$ – Joffan Dec 17 '17 at 21:07
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    $\begingroup$ @Abcd The reason you obtained the wrong answer is that you used only permutations. You have to choose the positions of the vowels, which requires the use of combinations, before you can permute the vowels and the consonants. $\endgroup$ – N. F. Taussig Dec 17 '17 at 21:16

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