SOLUTION: Seven cards, each marked a letter can be arranged to spell NUMBERS. how many three letter code words can be formed with 2 consonants and one vowel?
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Question 919672: Seven cards, each marked a letter can be arranged to spell NUMBERS. how many three letter code words can be formed with 2 consonants and one vowel? Found 2 solutions by stanbon, Theo:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Seven cards, each marked a letter can be arranged to spell NUMBERS. how many three letter code words can be formed with 2 consonants and one vowel?
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Assuming no repetition of letters::
Ans: 21C2*5C1 = 210*5 = 1050
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Cheers,
Stan H.
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5c2 is equal to 10
2c1 is equal to 2
3! is equal to 6
that would give you 120 possible 3 letter code words that could be generated where two of the letters would be consonants and one of the letters would be a vowel in each 3 letter combination.
the 5 consonants are:
nmbrs
the ten 2 letter combinations of consonants that could be generated by the 5c2 formula would be:
nm
nb
nr
ns
mb
mr
ms
br
bs
rs
the two 1 letter combination of vowels that would be generated by the 2c1 formula would be:
u
e
when you multiply 5c2 by 2c1, you get 10 combinations of consonants times 2 combinations of vowels to equal a total of 20 possible combinations where you have 2 consonants and 1 vowel in each combination.
a couple of those possible combinations would be:
nm1
nm2
nb1
nb2
etc.
each one of these 3 letter combinations, however, can generate 3! different codes.
for example:
nm1 can generate the following codes which are all different from each other because the order of the letters is different.
nm1
n1m
mn1
m1n
1mn
1nm
since this can occur with each of the 20 combinations given by 5c2 * 2c1, then you get 5c2 * 2c1 * 3! which is equal to 10 * 2 * 6 which is equal to 120 different code words.
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