The word EQUATION contains all five vowels. How many $3$ letter "words" consisting of at least $1$ vowel and $1$ consonant can be made from the letters of EQUATION?
Hi, would anyone be able to check the answer for this? The answers say 540, but I keep getting 270.
My means of working out were taking two cases: one with two vowels and one consonant, and one with one vowel and two consonants.
Thanks
On
I think your initial answer was correct and the "answer" provided to you was wrong.
Think how many different permutations of three letters we can select from a set of eight letters without replacement. It is $8 \cdot 7 \cdot 6 = 336$. (Right away this tells us the answer is wrong if you select without replacement, because the desired set is a subset of this set of words. But the answer $540$ is just as wrong if you select with replacement, via a similar argument.)
Now how many can you make containing only vowels, if you have five vowels available? That's $5 \cdot 4 \cdot 3 = 60$.
Now how many can be made with the three consonants you have? That's $3 \cdot 2 \cdot 1 = 6$.
But all the three letter words either have one vowel and two consonants, two vowels and one consonant, all vowels, or all consonants. Hence the set you're looking for consists of all words except the ones that are all vowels or all consonants, and the number of words in the set is
$$336 - 60 - 6 = 270.$$
Here is another way - take all three letter words and deduct those containing just vowels or just consonants. This comes to $$8\cdot 7 \cdot 6-5\cdot 4 \cdot 3 - 3\cdot 2 \cdot 1=336-60-6=270$$
If letters are allowed to be repeated the number is $$8^3-5^3-3^3=512-125-27=360$$
Another way of counting is to count the number of possibilities for each pattern of vowels and consonants.
$VVC: 5\times 4 \times 3=60$
$VCV: 5\times 3 \times 4=60$
$VCC: 5\times 3 \times 2=30$
$CCV: 3\times 2\times 5=30$
$CVC: 3\times 5\times 2=30$
$CVV: 3\times 2\times 5=60$
This gives $270$
I've done this longhand for clarity