I. Atmost three
C(n,r)=(n−r)!r!n!n=14r≤3=C(14,0)+C(14,1)+C(14,2)+C(14,3)=1+14+91+364=470
II. An equal number of 0s and 1s. This means that there will be 7 1s and 7 0s.
n=14r=7=C(14,7)=3432
III. An odd number of 1s. This means that 1s can be 1,3,5,7,9,11,13.
=C(14,1)+C(14,3)+C(14,5)+C(14,7)+C(14,9)+C(14,11)+C(14,13)=14+364+2002+3432+2002+364+14=8192
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