Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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: $4!/(2!*2!)$

share|cite|improve this question
Your approach is good, but it actually counts sentences that contain at least one of "new" and "this". That is, $5^4 - 3^4$ is the number of 4-word sentences made out of the five words [I am new at this], other than the 4-word sentences made out of [I am at]. – ShreevatsaR Aug 1 '12 at 12:16
Repeating "new" and "this" is allowed though. The only problem I currently see with this is that if I get "new this new this" I am counting it twice? or am I missing something? – user1411893 Aug 1 '12 at 12:30
@user1411893, you are also counting "new new new new" and "new I I I" etc. – picakhu Aug 1 '12 at 12:46
+1 for showing your thoughts. It allows much better answers. – Ross Millikan Aug 1 '12 at 13:35
up vote 2 down vote accepted

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

share|cite|improve this answer
Another way of reasoning to get to the same result would be: We can do $5^4$ sentences. However we have to remove all those sentences not containing "new", those are $4^4$. We also have to remove those not containing "this", again $4^4$. However now we have removed twice those containing neither "new" nor "this", so we have to add them once again. There are $3^4$ such sentences. Thus we have $5^4-2\cdot4^4+3^4$. – celtschk Aug 1 '12 at 13:44

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 $$ \binom{4}{2,2} $$



share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.