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How many 4-letter words with at least one consonant can be constructed from the letters A,B, C, D ,  and E ? (Note that B, C, and D are consonants, any word is valid, not just English language words, and letters may be used more than once.)

tertre  Feb 12, 2018
 #1
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Assuming that we can have "non-sense" words

 

We could have  a word with an "A"  and this letter could appear in any of 4 positions......and we could choose any of the other three consonants to occupy a second positon, then any two of the consonants to occupy a third position, and the last consonant would appear in the last position by default.......so we have

 

4 * 3 * 2 *1  =      24 words using just the A  and the other three consonants

 

The same result would occur if we just used the "E" and the other three consonants

 

If we used the "A"  and the "E," we would have 24 words using any two of the other consonants......but, we could choose any 2 of the 3 remaining consonants.....so   the total "words"   would be  24 * (3C2)  = 24 * 3  =  72

 

So....the total number of words would  be 24 + 24 + 72  =  120 "words"

 

winkwinkwink

lynx7  Feb 12, 2018
 #2
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Sorry, but that is wrong!

tertre  Feb 13, 2018
 #3
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lynx7, please don't copy an answer without giving credit.

 

If you find the answer on the forum already, just put a link to it...like this...

 

https://web2.0calc.com/questions/1-how-many-4-letter-words-with-at-least-one-vowel-can-be-constructed-from-the-letters-a-b-c-d-and-e#r1

 

( BTW this isn't the same question. )

hectictar  Feb 13, 2018
 #4
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Hi

this is a bit of a mind bender but as usual there is an easier path to the solution. i hope i have understood the question and interpreted it correctly. anyway here is my solution for what it is worth.

 

instead of working out the direct way of going through one consonant then two etc. i tried the "complement" method ie find the number of "words" with no restrictions and then the number of "words" with no consonants and subtract them.

 

number of "words" with no restrictions : 54  =  625  ie first position can have 5 letters as repetition is allowed so can all four positions.

 

number of "words" with no consonants (only vowels) : 24  =  16

 

So number of "words" with at least one consonant is 625 - 16 = 609

Guest Feb 13, 2018
 #5
avatar+86944 
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Wow!!... thanks, guest....that sure seems easier   !!!

 

 

cool cool cool

CPhill  Feb 13, 2018

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