Combinations using choose function (8 boys and 5 girls)

jullunety

New member
Joined
Jan 14, 2018
Messages
3
Question:
Among Tom's friends, there are 8 boys and 5 girls. In how many ways can he invite 2 of the boys and 3 of the girls to a party?


I am new to probability and I know that I can use the multiplication rule or the choose function, but both methods give me different answers.

Multiplication rule
Since there are only 2 boys he can choose so I would do this
(2 x 1 x 3 x 2 x 1) = 12 ways

Choose function

(13) (11)
( 2 ) x ( 3) = 12870 ways. or (13C2) x (11C3) = 12870 ways

Am I doing it all wrong? If so, how would you solve this problem through these 2 methods? Thanks in advance. :D
 
Last edited:
Question:
Among Tom's friends, there are 8 boys and 5 girls. In how many ways can he invite 2 of the boys and 3 of the girls to a party?


I am new to probability and I know that I can use the multiplication rule or the choose function, but both methods give me different answers.

Multiplication rule
Since there are only 2 boys he can choose so I would do this
(2 x 1 x 3 x 2 x 1) = 12 ways
What you've done here is to count permutations, not combinations. This is the number of ways he can order 2 boys, then 3 girls, who have already been chosen! You didn't even use the total numbers of boys and girls.

If you were not allowed to use the combination formula, you might say there are 8x7 ways to choose the boys, 5x4x3 ways to choose the girls, and then you'd have to divide by the number of different orders in which the 5 who are chosen could stand in a row, each of which counts as the same combination. I wouldn't bother trying to do it that way.
Choose function
(13) (11)
( 2 ) x ( 3) = 12870 ways. or (13C2) x (11C3) = 12870 ways

Am I doing it all wrong? If so, how would you solve this problem through these 2 methods?

This is closer to the question that was asked, but not quite. Here you correctly see that order doesn't count; he chooses a combination of girls and boys, not a permutation. The multiplication rule applies because he can choose boys first, then girls, without affecting the order in which individuals stand.

What you missed this time is that the 2 boys must be chosen from among the 8 boys present! He is not just choosing 2 of the 13 kids and calling them boys! Again, you didn't use the fact that there are 8 boys and 5 girls.

Try again, being sure to calculate

ways to choose 2 of 8 boys x ways to choose 3 of 5 girls.
 
What you've done here is to count permutations, not combinations. This is the number of ways he can order 2 boys, then 3 girls, who have already been chosen! You didn't even use the total numbers of boys and girls.

If you were not allowed to use the combination formula, you might say there are 8x7 ways to choose the boys, 5x4x3 ways to choose the girls, and then you'd have to divide by the number of different orders in which the 5 who are chosen could stand in a row, each of which counts as the same combination. I wouldn't bother trying to do it that way.


This is closer to the question that was asked, but not quite. Here you correctly see that order doesn't count; he chooses a combination of girls and boys, not a permutation. The multiplication rule applies because he can choose boys first, then girls, without affecting the order in which individuals stand.

What you missed this time is that the 2 boys must be chosen from among the 8 boys present! He is not just choosing 2 of the 13 kids and calling them boys! Again, you didn't use the fact that there are 8 boys and 5 girls.

Try again, being sure to calculate
ways to choose 2 of 8 boys x ways to choose 3 of 5 girls.


Thanks for taking the time to explain. I understand it better now. From what I can gather, this is how it should be done. (below)

Multiplication rule:
8 x 7 x 5 x 4 x 3 = 3360 (This is the number of permutations right?)
3360 / 2! x 3! = 280 combinations


Choose :
(8) (5)
(2) x (3) = 280.
 
Thanks for taking the time to explain. I understand it better now. From what I can gather, this is how it should be done. (below)

Multiplication rule:
8 x 7 x 5 x 4 x 3 = 3360 (This is the number of permutations right?)
3360 / 2! x 3! = 280 combinations


Choose :
(8) (5)
(2) x (3) = 280.

Permutations aren't relevant here, because the ordering of the boys and girls doesn't matter. (Picking Jim, then Bob, gives you the same result as picking Bob, then Jim).

You need to answer two questions

1) How many different ways are there to *choose* 2 boys out of the 8?

2) How many different ways are there to *choose* 3 girls out of the five?
 
Thanks for taking the time to explain. I understand it better now. From what I can gather, this is how it should be done. (below)

Multiplication rule:
8 x 7 x 5 x 4 x 3 = 3360 (This is the number of permutations right?)

3360 / 2! x 3! = 280 combinations


Choose :
(8) (5)
(2) x (3) = 280.
If you want possible permutations of choosing two boys from eight, there are 8 possibilities for the first choice and and 7 for the second = 8 * 7 = 56.

And the permutations on choosing three girls from five is 5 possibilities for the first one, 4 for the second, and 3 for the third = 5 * 4 * 3 = 60.

And 60 * 56 = 3360.

But as has been said, permutations are not relevant for this problem. Alice, Betty, and Claire is the same as Alice, Claire, and Betty or Betty, Alice, Claire, or Betty, Claire, Alice or Claire, Alice, Betty, or Claire, Betty, Alice. So you have to figure out the proper divisor for 3360, which you did without error. As the numbers get bigger, however, finding the proper divisor is more and more likely to lead to errors. So if you don't need permutations, don't use them. If you want combinations, use combinations. Less chance of error.

\(\displaystyle \dbinom {8}{2} * \dbinom{5}{3} = \dfrac{8!}{2! * (8 - 2)!} * \dfrac{5!}{3! * (5 - 3)!} = \dfrac{8 * 7}{2} * \dfrac{5 * 4}{2} = 28 * 10 = 280.\)
 
Last edited:
I get how to solve this problem by counting the possibilities, but IDK how to do it without actually needing to count the possibilities.

I think it would be 7+6+5+4+3+2+1 as the number of times you can pick 2 things out of 8

I think it would be 6+5+4+3+2+1 as the number of times you can pick 3 things out of 5 but that doesn't sit right with me, it seems too high.
 
I get how to solve this problem by counting the possibilities, but IDK how to do it without actually needing to count the possibilities.

I think it would be 7+6+5+4+3+2+1 as the number of times you can pick 2 things out of 8

I think it would be 6+5+4+3+2+1 as the number of times you can pick 3 things out of 5 but that doesn't sit right with me, it seems too high.
The number of ways to create a set of three items from five is 10. Your instinct was right.

Let's start with simple examples. We have a set of three distinguishable things. How many different sets of one of them can we have? 3. A or B or C. How many different sets of two of them can we have? 3. A, B or A,C or B, C. But what about B,A for example? We are not interested in order. A,B and B,A have the same elements and so are the same set. How many different sets of three can we have? 1, obviously.

OK let's try a set of two drawn from a set of eight distinguishable items. So A, B, C, D, E, F, G, and H are eight distinguishable items. We could have A, B or A, C or A, D, or A, E or A, F or A, G or A, H or B, C, or B, D, or B, E or B, F, or B, G or B, H or C, D or C, E or C, F or C, G or C, H or D, E or D, F or D, G or D, H or E, F or E, G, or E, H or F, G or F, H or G, H. That is 28 in total.

Now if if you want you could try four from ten distinguishable items through that method. As the numbers get bigger, this method of total enumeration becomes time consuming and very error prone. It is easier to remember a formula.

We want to know how many distinct ways we can draw k items from a set of n distinguishable items.

The answer is \(\displaystyle \dbinom{n}{k} = \dfrac{n!}{k! * (n - k)!}.\)

We define \(\displaystyle 0! = 1 \text { and } m \in \mathbb N^+ \implies m! = m * (m - 1)!.\)

So 0! is 1, 1! is 1, 2! is 2, 3! is 6, 4! is 24, and so on.

Let's use that formula on our examples.

One from three.

\(\displaystyle \dbinom{3}{1} = \dfrac{3!}{1! * (3 - 1)!} = \dfrac{3 * 2!}{1 * 2!} = 3.\)

That agrees with our method of total enumeration.

Two from three.

\(\displaystyle \dbinom{3}{2} = \dfrac{3!}{2! * (3 - 2)!} = \dfrac{3 * 2!}{2! * 1!} = \dfrac{3}{1} = 3.\)

That agrees. How about three from three?

\(\displaystyle \dbinom{3}{3} = \dfrac{3!}{3! * (3 - 3)!} = \dfrac{1}{0!} = \dfrac{1}{1} = 1.\)

That agrees. How about two from eight?

\(\displaystyle \dbinom{8}{2} = \dfrac{8!}{2! * (8 - 2)!} = \dfrac{8 * 7!}{2! * 6!} = \dfrac{8 * 7 * 6!}{2! * 6!} = \dfrac{8 * 7}{2} = 28.\)

Seems like a pretty neat formula.
 
Top