# Counting techniques

I&amp;#039;m not quite understand how to solve counting questions. Please could anyone shows me what the procedures are in solving these kinds of counting questions.

1 a.) Find the number of ways of arranging 6 girls and 3 boys to stand in a row so that all three boys are standing together.

b.) Find the number of ways of arranging 6 girls and 3 boys in a row so that no teo boys are standing next to one another.

c.)A group pf 12 people with 6 married couples is arranged at random in a line for photograph. Find the probabilty that each wife is standing next to his husband.

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(a) Consider the 3 boys as one group standing together, this together with the 6 girls make up 7 items. The number of ways to arrange 7 items = 7!

There are 3! arrangements of the boys within the group. So there are in total (7!)(3!) = 30,240 ways.

(b) Consider the 6 girls (G): pGPGPGPGPGPGp, there are 5 positions (P) between 2 girls and 2 positions (p) on the two ends. We need to put the 3 boys into these 7 possible positions. There are in total 7C3 ways to do it. So there are 7C3 = 35 ways for such arrangement.

(c) Consider each couple as a group standing together. There are 6! ways to arrange the 6 groups. Each group has two ways of standing husband left or wife left. So total number of ways = (6!)*(2^6) = 46,080

2009-12-14 23:16:35 補充：

I made a mistake in part (b), I have to include the arrangements of the boys 3! and girls 6!. So the answer is 35*3!*6! = 151200