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I have attached a screenshot of the proof of Pascal's rule and highlighted parts I don't understand.

enter image description here

1) why $(n-(k-1))$ on numerator is without factorial? I understand, that $(n-(k-1))$ is lesser than n!, but if n is left with factorial then there will be two (n-(k-1)), wouldn't there? Or maybe it's because there is $(n-(k-1))$ both on denominator and numerator and the $!$ cancels out?

2) Second question regards k on the second expression of the sum ( it is highlighted in yellow). I guess the question is the same, because again there is no $!$ and the $k$ is only left so that later one we can get rid off $(k-1)!$?

3) The last question regards the $(n-k+1)!$ on denominator on third line. The question is again similar to 1st one, where is $(n-k+1)$ and $(n-k)$? How can we simply increase the base of a factorial? ( I am not sure about the notation here)

Thank you!

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  • $\begingroup$ Yes, thank you. $\endgroup$ – Ieva Brakmane Apr 3 at 10:22
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1) we devide and multiply by (n-(k-1)) = (n-k+1) and that is 1 more than (n-k) so that (n-k)!(n-k+1) becomes (n-k+1)!. For example 3!=1*2*3=6, 4! =1*2*3*4=(1*2*3)*4=3!*4=24. Here they use the same principle.

2) they use the same principle as 1 to make k! From (k-1)!.

3) in the last highlighted part you have (n-(k-1))! Where you distrubute the - so that it becomes (n-k+1)! Both are equal to each other

Hope i have answered your questions properly, if not, just ask for further clarification on the parts you still don't understand!

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  • $\begingroup$ 2) But why they don't simply cancel the k! ? 3) Ok, yes, that was silly... $\endgroup$ – Ieva Brakmane Apr 3 at 10:44
  • $\begingroup$ So what they're doing is both deviding and multiplying with k. Why? Well we're aiming to have the same denominator so we can add the fractions together. As seen in the left fraction, the denominator is k!(n-k)!(n-(k-1)) which is also k!(n-k+1)!. We can see thqt we already have the (n-k+1)! But we don't have k!. What we do have is (k-1)! Which is one les than k. We use the same principle as in 1) so we multiply (k-1)! With k to get k!. Now both the denominators are equal so we can simply add the numerators together. Now it's just rewriting the numerator and we get the end result. $\endgroup$ – Viktor Apr 3 at 11:19
  • $\begingroup$ Ok, now I understand, I assumed that the left fraction is multiplied by (n-(k-1))! and second fraction multiplied by k! to partly equalize denominators. $\endgroup$ – Ieva Brakmane Apr 3 at 11:38

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