Arrange distinct groups of balls We have 10 green balls and 7 white balls. What should be estimated, is the number of the ways they can be arranged if those balls are distinct with no consecutive white balls.
Is my approach, that follows, correct? If not, how can it be solved?
Thank you.
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My approach:
There are 7! ways to arrange the 7 white balls in 7 "fixed" places.
Then, we have to find the number of ways the remaining 10 green balls can be arranged:
_W1_W2_W3_W4_W5_W6_W7_
So, there are 6 places that should hold at least 1 green ball:
_W1_G1_W2_G2_W3_G3_W4_G4_W5_G5_W6_G6_W7_
At this point, the no. of ways for this arrangment, is:
7! * 10 * 9 * 8 * 7 * 6 * 5
Now, there are 4 remaining green balls and 14 possible places that can hold any number of balls.
Finally, there are:
7! * 10 * 9 * 8 * 7 * 6 * 5 * C(17, 13) arrangements,
using the stars and bars approach for the last 4 green balls.

There are $10!$ ways to arrange the greens. Any such way leaves $11$ gaps (including the endgaps) into which to insert the whites, one white to each chosen gap. We can choose the gaps used in $\binom{11}{7}$ ways. Finally, multiply by $7!$. We get $10!\binom{11}{7}7!$.
Another way: The following is much closer to your approach, but does some correcting. Line up the   $7$ whites. Make $10$ "reserved for green" signs, and place one in each gap between whites. That leaves $4$ reserved signs. By Stars and Bars  they can be placed in the $8$ gaps (including endgaps) in $\binom{11}{4}=\binom{11}{7}$ ways. Finally, replace the reserved signs by greens. This can be done in $10!$ ways.
