1. How many 4-letter words with at least one vowel can be constructed from the letters A,B,C,D, and E?

4. In how many ways can 7 people be seated in a row of chairs if two of the people, Wilma and Paul, refuse to sit next to each other?

[ I learned  this trick from Melody......]

Let us consider the number of arrangements where Wilma  and Paul are sitting together

We can seat them in any of 6 positions and arrange the other people in 5! ways......but.....we can also arrange Wilma and Paul in two different ways in each of the 6 positions.....so.......the total arrangements where they sit together = 6 * 2 * 5!  =  1440 ways

And the total number of ways to seat all 7 people = 7!  = 5040 ways

So.......the total number of ways to seat all 7 people  minus the arrangements where Wilma and Paul sit togeher  = the number of arrangements where they don't sit together

5040 - 1440  =  3600 arrangements where they are not sitting together.......

1. How many 4-letter words with at least one vowel can be constructed from the letters A,B,C,D, and E?

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"

There are  99,999 - 10,000 + 1  = 90,000 five digit numbers

Let's just consider how many of them have no zeroes.......

We have 9 ways to select a non-zero digit for each of the positions, so this must be  9^5  = 59049  five digit numbers that have no zeroes........then  90,000 - 59,049   = 30,951 of them must contain at least one zero

Each hat matches up with 6 shirts, so ((1 hat)*(6 shirts)*(6 hats)) gives 36 shirt/hat combos. Each shirt/hat combo matches with 5 pants, so in total, there are 180 combos. There have to be 5 of one so that none have all one color. Or, 180=6!/4

3. I have 6 shirts, 6 pairs of pants, and 6 hats. Each item comes in the same 6 colors ( so that I have 1 item of each color). I refuse to wear an outfit in which all 3 items are the same color. How many choices for outfits do I have?

Assuming that two of the items can be the same color.......

We have  6*6*6   = 216 possible outfits if we can wear any color that we'd like........however....only 6 of these are all the same color......

So.....the total number of outfits you could wear, but avoiding all the same colors  =  216 - 6  = 210