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In how many ways 4 letters can be posted in two post boxes?
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Answer Posted / binesh
I think the robbins way
is right. But i couldnt
under stand why did
he multiply the result
by 2
the chances are
none in the fist box
and all in the other
box=4C0=1
1 in fist and 3 in
second=4C1=4
2 in first and 2 in
second=4C2=6
3 in first and 1 in
second=4C3=4
4 in fist and none in
second=4C4=1
so net chance is
4C0+4C1+4C2+4C3+4C4=16
| Is This Answer Correct ? | 4 Yes | 2 No |
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