Combinations Problem - Confirming My Solution

In summary, the conversation discussed the possibility of selecting four people from a committee of 12 for a function. The first question asked how many different groups of four were possible, which was determined to be 495. The second question asked for the number of groups in which either the chairperson or vice-chairperson must be included, but not both. This was calculated to be 450 using two different methods. However, the given solution was 240, which the individual is unsure about.
  • #1
Sean1
5
0
Hi everyone this is my first post. Please let me know if I am not following the proper etiquette.

I am helping out a friend with a problem. My answer to part (b) does not match the solution my friend has been given.

Four people are selected from a committee of 12 represent the committee at a particular function.
a) How many different groups of four are possible?
12C4 = 495
b) How many groups of four are possible of either the chairperson or the vice chairperson, but not both, must be in the group.
12C4 – 2C2 x 10C2 = 450
Alternatively
1C1 x 1C0 x 10C3 + 1C0 x 1C1 x 10C3 + 1C0 x 1C0 x 10C4
120 + 120 +210 = 450
I think this is correct, but the solution I have been given is 240?
 
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  • #2
For both the chairperson and vice-chairperson, we find there are 10 remaining people from which to choose for the other 3 spots on the committee, so we find:

\(\displaystyle N=2{10 \choose 3}=240\)
 
  • #3
MarkFL said:
For both the chairperson and vice-chairperson, we find there are 10 remaining people from which to choose for the other 3 spots on the committee, so we find:

\(\displaystyle N=2{10 \choose 3}=240\)

Many thanks! :cool:
 

FAQ: Combinations Problem - Confirming My Solution

What is a combinations problem?

A combinations problem is a mathematical problem that involves determining the number of ways to choose a subset of items from a larger set without regard to the order in which the items are chosen.

How do I solve a combinations problem?

To solve a combinations problem, you can use the formula nCr = n! / r!(n-r)!, where n is the number of items in the larger set and r is the number of items in the subset. Alternatively, you can use a combination calculator or list out all possible combinations and count them.

What is the difference between a combinations problem and a permutations problem?

A combinations problem involves choosing a subset of items from a larger set without regard to the order in which they are chosen, while a permutations problem involves choosing a subset of items from a larger set with regard to the order in which they are chosen.

Can I use combinations to solve real-world problems?

Yes, combinations can be used to solve real-world problems such as determining the number of possible outcomes in a lottery or the number of ways to select a team from a pool of candidates.

How do I confirm that my solution to a combinations problem is correct?

You can confirm your solution by checking your work with a combination calculator or by using the formula nCr = n! / r!(n-r)!. Additionally, you can double-check your work by listing out all possible combinations and counting them.

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