n! C(n, r) = r!(n – r)! Here we have n = 13 andr = 2. Substitute these values into the formula and simplify. n! C(n, r) = r!(n – r)! 13 ! C(13, 2) 2!(13 – 2)! In other words, there are x possible ways to pick 2 of the 13 denominations for the pairs.

Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
Chapter8: Areas Of Polygons And Circles
Section8.5: More Area Relationships In The Circle
Problem 31E
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n!
C(n, r) =
r!(n – r)!
Here we have n = 13 andr = 2. Substitute these values into the formula and simplify.
n!
C(n, r)
r!(n – r)!
13
C(13, 2) =
2!(13 – 2)!
=
In other words, there are
x possible ways to pick 2 of the 13 denominations for the pairs.
Transcribed Image Text:n! C(n, r) = r!(n – r)! Here we have n = 13 andr = 2. Substitute these values into the formula and simplify. n! C(n, r) r!(n – r)! 13 C(13, 2) = 2!(13 – 2)! = In other words, there are x possible ways to pick 2 of the 13 denominations for the pairs.
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