SOLUTION: #1. Prove for all integers n, k, and r with n ≥ k ≥ r that nCk×kCr = nCr×(n-r)C(k-r) #2. The binomial theorem states that for any real numbers a and b, (a + b)n =�

Question 1205092: #1.
Prove for all integers n, k, and r with n ≥ k ≥ r that nCk×kCr = nCr×(n-r)C(k-r)

#2.
The binomial theorem states that for any real numbers a and b,
(a + b)n =∑_(k=0)^n▒〖(n¦k) a^(n-k) b^k 〗 for any integer n ≥ 0.
Use this theorem to show that for any integer n ≥ 0, ∑_(k=0)^n▒〖〖(-1)〗^k (n¦k) 3^(n-k) 2^k 〗 = 1.

Found 2 solutions by mccravyedwin, math_tutor2020:
Answer by mccravyedwin(407) (Show Source): You can put this solution on YOUR website!

Those are the same except for the order of factors in the denominator. Your notation in the second one is incompatible with this site, making it so garbled, I can't make it out. Edwin

Answer by math_tutor2020(3816) (Show Source): You can put this solution on YOUR website!

Tutor mccravyedwin has covered problem 1.
I'll take a look at problem 2.

The binomial theorem where n is an integer and

Plug in a = 3 and b = -2

where

The nCk refers to the combination formula

The nCk values are found in Pascal's Triangle.
The nCk replaces the notation (n¦k) which is often written as

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