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So i have this specific question and i really can't seem to wrap my head around it. The question is as below;

There are 4 girls in a class of 16 students where 2 of the girls are twins. If we want to split them into two equal groups how many different ways can we do so for

a) when there is no restriction,

In this case i think its just 16C8

b) all girls to be in the same group

for this i think its 4C4 x 12C4

c) not wanting the twins in the same group,

This one im not too sure how to go about it.

I've always had trouble with understanding permutations and combinations and this one really got me thinking, i would appreciate any explanation i could get the answer doesn't really concern me but if i can finally understand what's happening here i would be more than grateful.

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  • $\begingroup$ I think you have (a) and (b) correct. For (c), build the two groups by placing the twins first (one in each group), and then arrange the remaining 14 people to complete the groups. $\endgroup$ Jan 18 at 13:32
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    $\begingroup$ @JaapScherphuis The groups are not labeled, so the answer to (a) is not correct. $\endgroup$ Jan 18 at 13:39
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Your a) is nearly correct, $\binom {16}{8}$ chooses the children to go in one group but the remaining children can also be selected in the same way; so you can get the same division twice. You can adjust for this by dividing by two (correct), $\frac 12 \binom {16}{8}$, or by picking an arbitrary child as being on one team and choosing the other $7$ chlldren to join her, $\binom {15}7$

Your b) is correct but a little more complicated than it needs to be: $\binom 44 = 1$ ("the girls are all on one team"), so the answer calculates as expected to be just $\binom {12}4$ ("choose the $4$ boys that are on the same team with the girls"). Here the group with the girls is thus labelled and there is no ambiguity about the choice as previously.

Solution to c) can start by allocating the twins one to each group (which are now therefore distinct eg. "Brenda's team" and "Glenda's team") and picking the other $7$ children to be on one of these.

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There are $4$ girls in a class of $16$ students where $2$ of the girls are twins. If we want to split them into two equal groups, how many different ways can we do so when there is no restriction?

You have counted each group twice, once when you select the group and once when you select its complement since choosing $8$ of the $16$ students to be in one group also determines which $8$ students are in the other group. The answer should be $$\frac{1}{2}\binom{16}{8}$$

Another way to see this is to suppose James is one of the members. We must choose which $7$ of the other $15$ students must be in the same group as James. Then the other eight students must be in the other group. Hence, there are $$\binom{15}{7}$$ ways to divide the students into two groups of equal size.

As you can verify, $$\frac{1}{2}\binom{16}{8} = \binom{15}{7}$$

There are $4$ girls in a class of $16$ students where $2$ of the girls are twins. If we want to split them into two equal groups, how many different ways can we do so when all the girls are placed in the same group?

Your answer is correct.

There are $4$ girls in a class of $16$ students where $2$ of the girls are twins. If we want to split them into two equal groups, how many different ways can we do so when the twins are not placed in the same group.

Suppose Katharine is one of the twins. Since her twin cannot be in the same group, we must choose $7$ of the other $14$ students to be in the same group as Katharine, which can be done in $$\binom{14}{7}$$ ways. The remaining students must in the group with Katharine's twin.

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