The five girls, named G1, G2,…G5 arranged the round-table sitting so that between each two of them there were at least two out of 12 boys , B1, B2,…B12.
In how many ways is such arrangement possible?
wlog, put G1 at the top of a circle.
There are 4! = 24 ways the remaining girls can be arranged.
Now we have 5 gaps into which we fit 12 boys.
Method "4in1gap" there are 4 boys in one gap, 2 in the others:
Combination(5,1) = 5 ways
Method "3in2gap" there are 3 boys in two gaps, 2 in the others:
Combination(5,2) = 10 ways
The patterns are thus 24 * 5 * 10 = 960 arrangements.
[Edit: Actually there are 2 errors here: I multiplied in my head by 4 instead of 5; and I forgot to include both the "4in1gap" and the "3in2gap" methods. The correct formula is:
24 * (5 + 10) * 10 = 3600 arrangements]
Now consider the 12! orders the boys could be in.
Final answer 3600 * 12! = 1,724,405,760,000 [corrected]
Edited on October 4, 2023, 8:12 am
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Posted by Larry
on 2023-10-02 15:19:16 |
Hopefully correct
