Probability

[Marierose]

How many 6 person committees are possible from a group of 10 people if there are no restrictions. Both Jim and mary must be on the committee? Either Jim or Mary but not both must be on the committee. Is there a formula to do this? If so please share!!!

 

[soroban]

Hello, Marierose!

How many 6-person committees are possible from a group of 10 people if there are no restrictions?

\(\displaystyle \text{There are: }\:_{10}C_6 \:=\:{10\choose6} \;=\;\frac{10!}{6!\,4!} \;=\; 210\text{ possible committees.}\)

If both Jim and Mary must be on the committee?

\(\displaystyle \text{Place Jim and Mary on the committee.}\)

\(\displaystyle \text{Then choose 4 more people from the remaining 8 people.}\)

\(\displaystyle \text{There are: }\:_8C_4 \;=\;{8\choose4} \;=\;\frac{8!}{4!\,4!} \;=\;70 \text{ ways.}\)

If either Jim or Mary( but not both) must be on the committee?

\(\displaystyle \text{Place Jim on the committee.}\)
\(\displaystyle \text{Then choose 5 people from the other 8 people (not including Mary):}\;\;{8\choose5} \:=\:56\text{ ways.}\)

\(\displaystyle \text{Place Mary on the committee.}\)
\(\displaystyle \text{Then choose 5 people from the other 8 people (not including Jim):}\;\;{8\choose5} \:=\:56\text{ ways.}\)

\(\displaystyle \text{Therefore, there are: }\:56 + 56 \;=\;112\text{ ways.}\)