Christmas Pictures!

For such a question,we use the gap method.

first place the girls.this can be done in 1 way as all girls are identical.

now there are 11 gaps between the girls(before the first girl and after the last girl included)

--G--G--G--G--G--G--G--G--G--G--

6 boys are to be placed in 11 gaps.

so there are 11c6*1!=462 ways

Assuming that each girl is identical and each boy is identical, we have to place 10 Gs and 6 Bs in a list (where G represents a girl and B represents a boy) so that no two Bs are next to each other. One way of doing this is to imagine each B as having a G to the right of it, so we treat BG as one block. In this way, no two Bs can ever end up next to each other. We then have the following 'blocks' to place:

BG

BG

BG

BG

BG

BG

G

G

G

G

However, of course this restriction also means that we will never end up with a B at the end of the list because we have made sure that there is always a G straight after each B. To solve this issue, we can introduce an 'imaginary' place at the end of the list and an extra girl. This means that now we can discard the last letter in any ordering of the list because it will always be a girl and then we will once again be left with a list of 6 boys and ten girls. This 'imaginary' place at the end of the list which gets discarded at the end allows us to end up with a boy in the last position because if the list ends with a BG block, the G will get discarded and we will end with a B.

So we now consider ordering the 6 'BG' blocks and the 5'G' blocks. There are 11 blocks in total and so there are 11C6 ways to place the 6 BG blocks. 11C6 = 462 ways so this is the answer.

The question should be reworded to clarify that the children are not all distinct because if they were taken to be distinct, the answer would be 462 x 10! x 6! to account for the different orderings within the boys and within the girls.

Note that the question asks you to find all arrangements such that all Boys and Girls are indistinguishable. This solution requires knowledge of Stars and Bars . If you aren't familiar with the Stars and Bars technique, I recommend that you read about it here: Stars and Bars

We will first arrange the Boys relative to each other in the line. There is only 1 way of arranging the Boys. This arrangement will simply look like this: BBBBBB

Next, to ensure that no two Boys are placed next to each other, we will put one Girl in between each pair of Boys. There is only 1 such arrangement: BGBGBGBGBGB

We still have 5 Girls remaining, and we have to slot them into our current arrangement of children. We can put the remaining Girls in 12 different spots - 10 spots in between any two of the children and two more spots at either ends. But, if we were to do this, we would have an over-lap of arrangements. There is in fact only one available spot in between two adjacent Boys. Thus, there are only 7 places where we could slot in the remaining girls. By Stars and Bars , we can calculate that there are 11C5 ways of slotting the girls into our arrangement.

Multiplying these values: 1 * 1 * 11C5 = 462