In how many ways 4 letters can be posted in two post boxes?

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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