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[scardizzle]

The word EQUATION contains all 5 vowels. How many 3 letter "words" consisting of at least 1 vowel and 1 consonant can be made from the letters of EQUATION?

okay, my working:

5C1(choosing 1 vowel from 5) x 3C1 (choosing one consonant)x 6C1(the rest) x 3!(no. of combs)

=540

and the answers seem to agree with me but if we work out the total no. of perms.

the answer would be 8 x 7 x6 = 336

which means my first answer cant be right... right?

Maybe im missing something...

 

[lychnobity]

The word EQUATION contains all 5 vowels. How many 3 letter "words" consisting of at least 1 vowel and 1 consonant can be made from the letters of EQUATION?

okay, my working:

5C1(choosing 1 vowel from 5) x 3C1 (choosing one consonant)x 6C1(the rest) x 3!(no. of combs)

=540

and the answer seem to agree with me but if we work out the total no. of perms.

the answer would be 8 x 7 x6 = 336

which means my first answer cant be right... right?

Maybe im missing something...

You're right, so where's the problem?

You're confusing me with the bolded part. 8 x 7 x 6 = ⁸P₃

⁸P₃ means choosing and arranging any 3 letters from EQUATION to form a word. This is wrong as it doesn't meet the conditions. eg EUA could be an arrangement, but this contains no consonants

 

[k02033]

I would do it as

all possible 3 letter words - all 3 letters are vowel- all 3 letter are consonants
P(8,3)- P(5,3)-P(3,3)

 

[scardizzle]

You're right, so where's the problem?

You're confusing me with the bolded part. 8 x 7 x 6 = ⁸P₃

⁸P₃ means choosing and arranging any 3 letters from EQUATION to form a word. This is wrong as it doesn't meet the conditions. eg EUA could be an arrangement, but this contains no consonants

how can a given condition have more permutations than without any restrictions?

I would do it as

all possible 3 letter words - all 3 letters are vowel- all 3 letter are consonants
P(8,3)- P(5,3)-P(3,3)

ah.. that' s a good way of thinking about it but what is the flaw in my logic?

 

[elmoateme]

I don't see how you got those results in the first place. The correct answer is 270.

Using 8P3 - 5P3 -3P3 the final answer was 270

Using a different method I got the same answer. Here's what I did, hope it helps.


Ways to choose 2 vowels & 1 consonant = 5C2 x 3C1 = 30
Ways to choose 1 vowel & 2 consonants = 5C1 x 3C2 = 15

Therefore total # of letter combinations = 45

Now, there are 3! ways to arrange each combinations once choosen.

Therefore Total # of words formed = 45 x 3! =45 x 6 =270 as above.

I'm not really sure why your why didn't work, but for all these questions you really need to take it step by step. Also, that way if you get it wrong you get marks for working. Hope I was of some help

 

[scardizzle]

ah... of course! thanks for the all help everyone