hy2000

Q1: nCr+1 - nCr-1 = n+1Cr-1 - n+1Cr


Q2: nC3 = 5nC1


Q3: nC1+ nC2 = 3n


I would like to know how to solve these questions with detail steps, thank you!!

My answer on Q2 is 1 and Q3 is 4 but the model answer are 7 and 5 respectively.

skipjack

(1) Is the equation typed correctly? It reduces eventually to n = 2r - 1, where r > 1.

(2) n(n - 1)(n - 2)/6 = 5n, so n((n - 1)(n - 2) - 30) = 0. Hence n = 0 or 7.

(3) 3n = nC1 + nC2 = n + n(n - 1)/2, so 0 = n(2 + n - 1 - 6). Hence n = 0 or 5.

hy2000

Question 1 should be

Prove LHS=RHS

skipjack

There must be an error in question (1), as the equation given isn't true in general.

Maybe its right-hand side should have had n+1Cr+1 instead of n+1Cr-1.

It then follows from nCr-1 + nCr = n+1Cr and nCr + nCr+1 = n+1Cr+1.

hy2000

Is it possible that you can prove question 1 has an error in detail steps, cause this question is from my textbook so I want to know what is wrong, thank you.

skipjack

(1) In its original form, the equation doesn't hold for n = 4 and r = 2, say, so if the third term is n+1Cr-1 in the textbook, the textbook is in error.

hy2000

Would you prove that

nC(r+1) - nC(r-1) = (n+1)C(r+1) - (n+1)Cr


Later I contact the publisher to see if there are any typo problem in this question, thanks a lot

topsquark

Can you at least give it a try yourself and show us what you've done?

-Dan

skipjack

I've already given two equations that are equivalent to Pascal's rule (also known as Pascal's identity), and which you can use without proof (but several proofs are given in the linked article). If you subtract one of the equations from the other, thus eliminating the nCr terms, the resulting equation can be rearranged to give the equation quoted above that your textbook probably intended.

hy2000

Sure I will try it myself, if I have any question I will ask you