How many 4 worded sentences can a list of 5 words make if two of them must be in that sentence? Suppose we have:
I am new at this - (5 words)
how many 4 worded sentences can we make with this if "new" and "this" must appear in the sentence.
I think its :
.# of sentences we can make with any word in it  - # of sentences we can make with no mention of "new" and "this" in them
so:
$5^4 - 3^4 
= 544$
Is that the right way of doing these type of questions?
Thanks
So we just got the solutions from the professor and it seems the answer was 150
He did it by saying:
"new" and "this" appear once: $3c2 *4!$
"new" appears twice, "this" once: $3c1 * 4!/2! $
"this" appears once, "new" twice: $3c1 * 4!/2! $
"this", "new" appear twice: $4!/(2!*2!)$
 A: ShreevatsaR is correct, so you need also to subtract those sentences with exactly one of "new" or "this" (perhaps multiple times).  How many are those?   Let's just do "new" and not "this"-the other will be the same by symmetry.  We have four words to choose from and must delete the ones that don't have any "new"s, so it is $4^4-3^4$.  The final result would then be $5^4-2\cdot 4^4 + 3^4$.  This is an example of the inclusion-exclusion principle
A: 5 different words (A, B, C, D, E)
A, B must appear. Repetition is allowed. How many permutations can you make with with 4 words.
We have to take cases.
Case 1. There is no repetition of A and B
A, B occupy 2 slots. Other 2 slots are free to be { C, D, E }.
Pick two slots for A, B and permutate, then number of functions from {C, D, E} -> {1, 2} (last 2 remaining slots)
$$
2\binom{4}{2}\cdot3^{2}
$$
Case 2. There is repetition of A but no repetition of B
Since the numbers are small, we can just write out the subcases explicitly. If not we'd have to use sigma notation with multinomials.
Subcase I. A repeated twice
We have a selection of A x 2, B x 1, and { C, D, E } x 1
$$
3\binom{4}{2,1,1}
$$
Subcase II. A repeated trice
$$
\binom{4}{1}
$$
Case 3. There repetition of A but no reptition of B
The same as case 2.
Case 4. There is repetition of both A and B
The only selection we have is A x 2, B x 2
$$
\binom{4}{2,2}
$$
Result
194
