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How many different words of length $5$ can be formed using all the letters m, a, t, h, s without repetition

If the first letter must be a vowel?

If the first letter must be a consonant?

It is a question from my book in the permutations chapter. I have no idea how to approach it. I know about factorials but I get confused when they specify the vowel and the consonant cases.

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  • $\begingroup$ Do all the letters have to be used in each "word"? (Also, I assume that the "words" can just be strings of letters, rather than needing to be actual English words.) $\endgroup$ – Michael Seifert Nov 29 '17 at 18:15
  • $\begingroup$ Yes any strings of letters $\endgroup$ – Deni Katsman Nov 29 '17 at 18:16
  • $\begingroup$ How long are the words supposed to be - are repetitions allowed? $\endgroup$ – Mark Bennet Nov 29 '17 at 18:20
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    $\begingroup$ 5 character long, no repetitions, sorry for not specifying $\endgroup$ – Deni Katsman Nov 29 '17 at 18:21
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If the first letter must be a vowel: you have the "a" as first letter and you can permutate 4 letters, thus you can create $4!$ words (meaningless I suppose)

If the first letter must be a consonant: you have 4 possibilities for the first letter and you can permutate the remaining 4 letters, thus you can create $4\cdot4!$ words

NOTE: as a check summing up the two results you obtain $$4!+4 \cdot 4!=5 \cdot 4!=5!$$ that is the total number of permutation for the 5 letters.

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  • $\begingroup$ thanks, and if it is a consonant? $\endgroup$ – Deni Katsman Nov 29 '17 at 18:17
  • $\begingroup$ I've completed the answer :) $\endgroup$ – user Nov 29 '17 at 18:18
  • $\begingroup$ Thank you very much, I WISH I could pull it out from the problem like you did. I literally did not even see this as an option. $\endgroup$ – Deni Katsman Nov 29 '17 at 18:19
  • $\begingroup$ You are welcome, you have only to practice, it wasn't a problem so hard :) $\endgroup$ – user Nov 29 '17 at 18:21
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I assume your book means permutations, not English words. If it means English words the question has a lot less to do with maths. To find the answer, notice that there are five letters in maths, so there are five different letters that can go in the first position, either m can come first, or a, or t, or h, or s. Next, you've used one of these letters, so only 4 are available to come second in the word. Now you've used 2 letters, so only 3 are available to go third. This continues, so the answer is 5x4x3x2x1=120. Now if you want to find the number of ways you can do it with consonant first, your question is more complicated. For vowels, it is easy, as there is only one letter that can come first: a. A is in spot one. There are four options for spot 2. 3 options for spot 3. 2 for 4, and 1 for 5. The answer is 4x3x2x1=24. For starting with a consonant, 4 letters can come first, because a can't, and now you only have 4 letters left, so only 4 letters for spot 2, 3 for 3, 2 for 4, and 1 for 5. The answer then is 4x4x3x2x1=96.

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  • $\begingroup$ Yeah that's what they mean I guess, but that's exactly how it was spelt in the book. Thank you very much for the explanation! $\endgroup$ – Deni Katsman Nov 29 '17 at 18:29

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