2
$\begingroup$

4 girls and 6 boys are lining up to receive trophies at a competition. How many ways are there to arrange them so that exactly two girls are next to each other?

My thoughts were to group two girls together, then place 2 boys between the 3 girl groups and distribute the remaining 4 boys in the 4 places adjacent to the girl groups, but I'm not sure how to accomplish this. I'm not certain of the answer either; any help is appreciated!

$\endgroup$
  • $\begingroup$ are the people in each gender distinguishable ? $\endgroup$ – user451844 Jul 25 '17 at 0:08
  • $\begingroup$ I believe that the intent of the question was that they are not distinguishable; apologies for leaving that out! $\endgroup$ – Elk Chu Jul 25 '17 at 0:33
3
$\begingroup$

If there are exactly $2$ girls next to each other, then we must distribute $1$ group of $2$ girls and $2$ groups of $1$ girl among the spaces between the $6$ boys. There are $7$ spaces where we can place the groups of girls: $$ \_b\_b\_b\_b\_b\_b\_ $$ So, there are $7$ ways to pick the placement of the $2$ girls next to each other and $\binom62$ ways to pick the placement of the $2$ individual girls for a total of $$ 7\cdot\binom62 = 7\cdot 15 = \boxed{105} $$ ways.

$\endgroup$
  • $\begingroup$ This solution is correct if all boys are indistinguishable and all girls are indistinguishable. The OP has to clarify is that's the assumption here, or if all persons are distinguishable from each other. $\endgroup$ – zipirovich Jul 25 '17 at 0:17
  • $\begingroup$ Good point, that was careless of me, thank you. $\endgroup$ – Taisuke Yasuda Jul 25 '17 at 0:18
  • 1
    $\begingroup$ No, you're fine! It's the OP who carelessly didn't provide some very important information. $\endgroup$ – zipirovich Jul 25 '17 at 0:19
  • $\begingroup$ Additionally, you should not do all the asker's work for them. Provide hints and help, but let them complete the assignment. $\endgroup$ – Graham Kemp Jul 25 '17 at 0:30
  • $\begingroup$ Thanks for your feedback, I will be careful of that! $\endgroup$ – Taisuke Yasuda Jul 25 '17 at 0:31
0
$\begingroup$

Since you have 4 girls and 4 boys, this means they have names(this usually assumed)! So you have total of 10 Persons. If they formed a line, it would n! possible outcomes. Since they have names and a line means it has a start and an end. Now you have to choose 2 girls this can be done in $\binom{4}{2}$ for example: Lisa+Anna or Anna+Lisa so we have to multiply this by 2 or 2!(2-Permuations) Okey now "glue" them together and you get a supergirl! This means you have 9 Persons of which one is supergirl. So the answer would be $$\binom{4}{2}*2!*9!$$

I have a problem now , i am counting also combination where the supergirl stand next to one or two two girls, so we have to subtract this. So we apply the procedure again we need to remove blocks of two 2-blocks of girls, 3-blocks and 4-blocks of girls and one has to be carefull, if you remove all blocks of 2-blocks you also remove the 4-block of girls. I think one has to proceed by inclusion exclusion formula from here.

Two 2-blocks of girls: Lets calculate how many possible combination we have if we have two block of girls. Choose 2 girls $\binom{4}{2}$ the others 2 form automaticaly a block, so we have two blocks X1 and X2 of supergirls. Each of then could be permutated in 2! so in total: $\binom{4}{2}*2!*2!$. Now we have two supergirls X1 and X2 and in total 8 Persons, and 8! to arrange them. This leads to $\binom{4}{2}*2!*2!*8!$ unwanted combinations. In this combination is included the block of 4 girls, X1X2 on any position.

One 3-blocks of girls: Choose 3 girls $\binom{4}{3}$,could be permutated in 3! so in total: $\binom{4}{3}*3!$. Now we have one supergirl X1 and one normal girl and in total 8 Persons, and 8! to arrange them. This leads to $\binom{4}{3}*3!*8!$ unwanted combinations. In this combination is included the block of 4 girls, X1X2 on any position.

And now one could subtract the unwanted combination from the wanted, but you have to add at some point the block of 4 girls.

Have i forgotten something?

This "glueing" solution is based on course:https://www.coursera.org/learn/modern-combinatorics/lecture/Dmnpx/piervokursniki-v-kinotieatrie Or Chapter 1 in Miklos Bona Introduction to enummerative combinatorics, theare excersises with solutions. The other possibility would be first you position the 6 boys and then use the "spaces in between solution" in the answer above. Its much faster in my opinion.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.