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No. of ways in which 5 boys and 5 girls can stand in a queue in such a way that exactly 4 girls stand consecutively in the queue?

The image of the question can be seen on the link below. It is google link as I can't upload image. Q44 of the pic is what am I am asking part 2.

My attempt:

Out of 5 girls 5C4 girls are chosen and can be permutated in 4! ways. The remaining can go in the remaining 6 places and can be permutated in 7! ways. But unfortunately the answer is not that.

Rakshit Jha
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    Do you mean $4$ particular girls or any $4$ of them? – Nitin Uniyal Dec 24 '16 at 05:31
  • What are your thoughts? What have you tried? Where are you stuck? You need to provide context for your question. Right now, it just looks like you want somebody to do your homework for you; that's not what this site is for. If you add some appropriate context, we will be happy to help. – Greg Martin Dec 24 '16 at 05:32

2 Answers2

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I assume you need exactly 4 girls together.

Hint 1: If you have 4 girls together, depending on where they are in the line you will need a boy on one side of them or the other or a boy on both sides. How many ways can you do this?

Answer 1: 5 ways with BGGGGB, 1 way with GGGGB, and 1 way with BGGGG. This means there are a total of $5+1+1=7$ ways to do this.

Hint 2: Once you have placed the 4 girls, where can the last girl go?

Answer 2: There 4 ways to place the last girl with BGGGGB and 5 ways for either GGGGB or BGGGG. Now there are $4\cdot 5+5\cdot 1+5\cdot 1=30\,\,$ ways to do all of this.

Hint 3: The first two hints should tell you how many different orderings of indistinguishable boys and girls there are (based on the work above, the answer is 30). So now if we fix an ordering of boys and girls, how many ways are there to place distinguishable boys and girls in the assigned positions?

Answer 3: There are $5!$ ways to place the girls and $5!$ to place the boys for a total of $$ (5!)^2\cdot 30=432,000. $$ Observe that $$ \binom{6}{2}\cdot\binom{5}{4}\cdot4!\cdot2!\cdot5!=\frac{6\cdot 5}{2}\cdot5\cdot4!\cdot2!\cdot5!=30\cdot (5!)^2=432,000. $$

Laars Helenius
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  • Out of 5 girls 5C4 girls are chosen and can be permutated in 4! ways. The remaining can go in the remaining 6 places and can be permutated in 7! ways. But unfortunately the answer is not that. – Rakshit Jha Dec 24 '16 at 06:25
  • No. You aren't following the hints correctly. Once an ordering of girls and boys are chosen there are $5!$ ways to place the boys in their spots and $5!$ ways to place the girls in there spots. – Laars Helenius Dec 24 '16 at 17:42
  • See my edits above! – Laars Helenius Dec 24 '16 at 18:11
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Case 1-

If particular girls.

Group 4 girls as 1.

So now we have total 5 boys + 1 girl + 1 group of girls = 7

These 7 can be arranged in 7! Ways.

Group of girls can be further arrange in 4! Ways.

So number of ways = 4! * 7!

Case 2-

If not particular girls.

Then first we pick them using C(5,4)

Now Group 4 girls as 1.

So now we have total 5 boys + 1 girl + 1 group of girls = 7

These 7 can be arranged in 7! Ways.

Group of girls can be further arrange in 4! Ways.

So number of ways = C(5,4) * 4! * 7!

Kanwaljit Singh
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