Pt1420 Unit 1 Assignment 1
1. Explain in words why the equation {n \choose k} = {n \choose n -k} is true.
The equation C(n;k) stands for "n choose k", and represents the number of k length element combinations which can be formed from a set of n elements (order doesn't matter). This equations also states that the result from the left side of the equations equals the number of combinations not chosen n-k on the right side.
This is true because the left side calculates the number of possible ways to create possible combinations for k chosen out of n objects. The left side does the exact same thing but accomplishes it by instead counting the number of those not selected n-k
If you look at a pascal triangle you will find the resulting combinations k on left side always match the number not selected n-k on the right side, so they are equivalent. …show more content…
This first problem although simple caused me nightmares when trying to provide a verbally explanation. The math behind it appeared fine, however I'm not totally satisfied with my description. If you could point me to a more concise explanation it would be greatly appreciated.
2. Use the definition {n \choose k} = \frac{n!}{k!(n -k)!} to show that the equation in question (1) is true.
This equations speaks to the relationship between combinatorials and permutations.
Proven {n \choose k} = \frac{n!}{k!(n -k)!}
P(n;k) = n! / (n-k)! permutation equation, (number of combinations (order doesn't matter) of k length combinations derived from a set of n elements
k! = number of combinations k elements can be recombined (same characters different orders).
therefore if you divide the number of permutations by k! you get
{n \choose k} = \frac{n!}{k!(n -k)!}
Prove {n \choose k} = {n \choose n -k}
The equation C(n;k) stands for "n choose k", and represents the number of k length element combinations which can be formed from a set of n elements (order doesn't matter). This equations also states that the result from the left side of the equations equals the number of combinations not chosen n-k on the right side.
This is true because the left side calculates the number of possible ways to create possible combinations for k chosen out of n objects. The left side does the exact same thing but accomplishes it by instead counting the number of those not selected n-k
If you look at a pascal triangle you will find the resulting combinations k on left side always match the number not selected n-k on the right side, so they are equivalent. …show more content…
This first problem although simple caused me nightmares when trying to provide a verbally explanation. The math behind it appeared fine, however I'm not totally satisfied with my description. If you could point me to a more concise explanation it would be greatly appreciated.
2. Use the definition {n \choose k} = \frac{n!}{k!(n -k)!} to show that the equation in question (1) is true.
This equations speaks to the relationship between combinatorials and permutations.
Proven {n \choose k} = \frac{n!}{k!(n -k)!}
P(n;k) = n! / (n-k)! permutation equation, (number of combinations (order doesn't matter) of k length combinations derived from a set of n elements
k! = number of combinations k elements can be recombined (same characters different orders).
therefore if you divide the number of permutations by k! you get
{n \choose k} = \frac{n!}{k!(n -k)!}
Prove {n \choose k} = {n \choose n -k}