A committee of #4# people be selected from a group of #7# people in #35# ways.
Explanation:
As the order of people does not matter, it is #\C_4^7#
i.e. #[7xx6xx5xx4]/[1xx2xx3xx4]=[7xxcancel6xx5xxcancel4]/[1xxcancel2xxcancel3xxcancel4]=35#
Hence, a committee of #4# people be selected from a group of #7# people in #35# ways.
Natalia P. How many ways can a 2-person subcommittee be selected from a committee of 7 people? I need help with this problem please More
1 Expert Answer
Omari S. answered • 11/24/15
Johns Hopkins Grad Student and MCPS Math Teacher w/ Eng. Background
Use the combination rule of n!/k![n-k]! where n is the number of options [7] and k is the number of slots [2].
Do this if the order does not matter.
7!/2![7-2]! = 7*6*5*4*3*2*1/2*1*[5*4*3*2*1] = 7*6/2 = 21
If, however, the order does matter, then use the permutation rule of n!/[n-k!] with n = 7 and k =2 giving 7*6 = 42.
Still looking for help? Get the right answer, fast.
OR
Find an Online Tutor Now
Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.
The first selection can be any one of 7 people.
The second one can be any one of the remaining 6.
The third one can be any one of the remaining 5.
So the subcommittee can be selected in [7 x 6 x 5] = 210different ways.
But ... there aren't that many different subcommittees that can be formed.
Whether you select ABC, ACB, BAC, BCA, CAB, or CBA, you still have the same 3 people
on the subcommittee. You can form the subcommittee in 210 different ways, but
each group of the same three people gets selected six times, in different ways.
So there are only [210/6] = 35 possible different subcommittees.