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 =âˆ

Algebra ->  Permutations -> 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 =∠     Log On


   

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(406) About Me  (Show Source):
You can put this solution on YOUR website!

%28nCk%29%28kCr%29+=+%28nCr%29%28n-r%29C%28k-r%29

%28n%21%2F%28k%21%28n-k%29%21%29%29%2A%28k%21%2F%28r%21%28k-r%29%21%29%29%22%22=%22%22

%22%22=%22%22

n%21%2F%28%28n-k%29%21r%21%28k-r%29%21%29%22%22=%22%22n%21%2F%28r%21%28k-r%29%21%28n-k%29%21%29%29

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) About Me  (Show Source):
You can put this solution on YOUR website!

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

%28a%2Bb%29%5En+=+sum%28nCk%2Aa%5E%28n-k%29%2Ab%5Ek%2Ck=0%2Cn%29 The binomial theorem where n is an integer and n+%3E=+0

%283%2B%28-2%29%29%5En+=+sum%28nCk%2A3%5E%28n-k%29%2A%28-2%29%5Ek%2Ck=0%2Cn%29 Plug in a = 3 and b = -2

%283-2%29%5En+=+sum%28nCk%2A3%5E%28n-k%29%2A%28-1%2A2%29%5Ek%2Ck=0%2Cn%29

%281%29%5En+=+sum%28nCk%2A3%5E%28n-k%29%2A%28-1%29%5Ek%2A2%5Ek%2Ck=0%2Cn%29

1+=+sum%28%28-1%29%5Ek%2AnCk%2A3%5E%28n-k%29%2A2%5Ek%2Ck=0%2Cn%29

sum%28%28-1%29%5Ek%2AnCk%2A3%5E%28n-k%29%2A2%5Ek%2Ck=0%2Cn%29+=+1 where n+%3E=+0



The nCk refers to the combination formula
nCk+=+%28n%21%29%2F%28k%21%2A%28n-k%29%21%29
The nCk values are found in Pascal's Triangle.
The nCk replaces the notation (n¦k) which is often written as %28matrix%282%2C1%2Cn%2Ck%29%29