# I I'm having trouble seeing the big picture of this proof.

1. Aug 10, 2016

### Terrell

I don't see how it proves that (n-r+1, r) is the number of r-combinations of X which contain no consecutive integers.

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2. Aug 10, 2016

### haruspex

Is it that you do not see that it is a bijection, or that establishment of the bijection proves the result?

3. Aug 10, 2016

### Terrell

i can't see that the establishment of the bijection proves the result. please do help me

4. Aug 10, 2016

### Terrell

after giving it some thought, i think i've got it. after applying the bijective function to set S, we can observe that the number of elements of the original set is equal to the number of elements in the set produced by the bijective funtion. however, the only difference is that the original set consists of non-consecutive integers and the "new" set consists of consecutive integers. since both of the sets contains n-r+1 elements, the number of ways to choose r-elements from the "new" set is (n-r+1, r) which should also equal for the original set. did i got that one right?

5. Aug 10, 2016

### haruspex

Yes.

6. Aug 11, 2016

### Zafa Pi

I think you were being too generous. When Terrell said, "original set consists of non-consecutive integers and the "new" set consists of consecutive integers.", he was referring to S and f(S) and that is in fact not necessarily true.

7. Aug 11, 2016

### haruspex

You are right, I missed that it said "consists of" instead of "may contain", but that is probably just a minor slip in expressing it.

8. Aug 11, 2016

Agreed.