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I found following Problem in Introductory Combinatorics text book:

How many eight-letter words can be constructed by using the 26 letters of the alphabet if each word contains three vowels? It is understood that there is no restriction on the number of times a letter can be used in a word

Self Attempt: I tried to solve the problem in following manners:
There are 5C3 ways we can select 3 vowels from the alphabet and 21C5 ways to select 5 consonant from 21 consonants. For each such 8 characters selection we can arrange the word in $8^8$ ways (as repetition of character is allowed).
My answer: $5C3 \cdot 21C5 \cdot 8^8$

But textbook has the following answer: $8C5 \cdot 5^3 \cdot 21^5$

Could you please help me know where I am going wrong.

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  • $\begingroup$ This tutorial explains how to typeset mathematics on this site. $\endgroup$ Feb 7, 2022 at 13:10
  • $\begingroup$ @N.F.Taussig Thanks. $\endgroup$
    – fluty
    Feb 7, 2022 at 13:11

1 Answer 1

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The textbook is saying "choose the $5$ out of $8$ positions for consonants (so $3$ for vowels), then choose the $3$ vowels possibly with repetition and the $5$ consonants possibly with repetition"

$^{5}C_3$ would be choosing three distinct vowels in any order rather than the $5^3$ since order matters and repetitions are allowed, and similarly $^{21}C_5$ should be $21^5$ for the same reason.

I am not clear what your $8^8$ was intended to be. If the question had prohibited repeated letters, the answer would have been $8!\,^{5}C_3\, ^{21}C_5$

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  • $\begingroup$ Have updated my question. It was a typing mistake. My final answer was 5C3 * 21C5 * 8^8 $\endgroup$
    – fluty
    Feb 7, 2022 at 13:06
  • $\begingroup$ Yes - I see. I have edited my answer $\endgroup$
    – Henry
    Feb 7, 2022 at 13:07
  • $\begingroup$ 8^8 means, There are 8 possible choices for each letter for each position in 8 letter word. Since there is no restriction on number of times a letter can be used, I think repetition is allowed. $\endgroup$
    – fluty
    Feb 7, 2022 at 13:13
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    $\begingroup$ @fluty $8^8$ would count for example the possibility of the same consonant being used $8$ times, which would preclude there being $3$ vowels, and many similar cases. Looked at another way, you have chosen $3$ distinct vowels and $5$ distinct consonants in your $\,^{5}C_3\, ^{21}C_5$ but if you do not use them all then how do know which ones were the unused ones you chose? $\endgroup$
    – Henry
    Feb 7, 2022 at 13:40
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    $\begingroup$ Thats because repetition is allowed $\endgroup$
    – fluty
    Feb 7, 2022 at 16:44

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