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Could you please explain writing it out how 12C3 ends up being 220. I know how to look it up on the calculator but I need to know how to show my work by writing it out fully

Could you please explain writing it out how 12C3 ends up being 220. I know how to look it up on the calculator but I need to know how to show my work by writing it out fully

College Algebra
College Algebra
10th Edition
ISBN:9781337282291
Author:Ron Larson
Publisher:Ron Larson
Chapter8: Sequences, Series,and Probability
Section8.7: Probability
Problem 6ECP: In Pennsylvania’s Cash 5 game, a player chooses five different numbers from 1 to 43. If these five...
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Hi,

Could you please explain writing it out how 12C3 ends up being 220. I know how to look it up on the calculator but I need to know how to show my work by writing it out fully. Thank you.

Answer to Problem 8TY The probability that all three jury members are men is 0.045. Explanation of Solution Given info: In a jury selection, there are 12 people in a group in which 5 peoples are men and seven are women. Also, three members are randomly selected. Calculation: Combination Rule: There are n different items and r number of selections and the selections are made without replacement, then the number of different combinations (order does not matter counts) is given by, n! nCr = (n-r)!r! The possible number of ways of choosing 3 men out of 5 men is, Substitute 5 for 'n' and 3 for 'r', then the number of possible different treatment groups is 5! 5C3= (5-3)13! = 10 The number of possible ways for choosing three people is, Substitute 5 for 'n' and 3 for 'r', then the number of possible different treatment groups is 12! 12C3 = (12-3)13! = 220
Transcribed Image Text:Answer to Problem 8TY The probability that all three jury members are men is 0.045. Explanation of Solution Given info: In a jury selection, there are 12 people in a group in which 5 peoples are men and seven are women. Also, three members are randomly selected. Calculation: Combination Rule: There are n different items and r number of selections and the selections are made without replacement, then the number of different combinations (order does not matter counts) is given by, n! nCr = (n-r)!r! The possible number of ways of choosing 3 men out of 5 men is, Substitute 5 for 'n' and 3 for 'r', then the number of possible different treatment groups is 5! 5C3= (5-3)13! = 10 The number of possible ways for choosing three people is, Substitute 5 for 'n' and 3 for 'r', then the number of possible different treatment groups is 12! 12C3 = (12-3)13! = 220
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