Help with this permutation & combination question please

There are 4 girls, 3 boys. How many ways to line them up if the boys have to be in order of increasing age. ( The boys don't have to be together)

easy way
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just place the 4 girls in any of 7 slots
for each of these,, the boys can be placed in only 1 way, so
# of ways = 7P4 = 7*6*5*4 = 840 <---------

Start by working out where thhe boys are.
Call the boys, in order of ascending age A, B and C and number the line positions from 1 to 7.

The highest position A can be in is number 5 (leaving B & C) in 6 & 7 - 1 combination
If A is in number 4, we can have B5C6 or B5 C7 or B6 C7 = 3 combinations
If A is in number 3, B can be anywhere from 4 to 6, C has respectively 3, 2 or 1 place to be = 3+ 2 + 1 = 6 combinations
If A is in 2, B has 3 to 6, gives 4 + 3 + 2 + 1 = 10 possibilities
If A is in 1, 15 possibilities

All up that gives us 1 + 3 + 6 + 10 + 15 = 35 arrangements of the boys.

For each of these, the 4 girls can be arranged in the remaining 4 spaces in any order: 4 * 3* 2 * 1 = 24 arrangements

All together, that gives us 35 * 24 = 840 arrangements

* I think Gaurav would be right if the boys were not allowed to be together (ie boys must be separated by a girl) but you have just said they need not be together.

first arrange the girls in 4! ways

now there are 5 places besides the girls
so select 3 places out of 5 in 5C3 ways

and arrange the boys in 1 way as they have to be in increeasing order of their ages

so answer is 240