a) We should choose 4 place from 14 for 1. And another will fill with 0.
"\\dbinom{14}{4}=\\frac {14!} {4!(14-4)!}=1001"
b) We should choose 4, 3, 2, 1, 0 place from 14 for 1. And another will fill with 0.
"\\dbinom{14}{0} + \\dbinom{14}{1} + \\dbinom{14}{2} + \\dbinom{14}{3} + \\dbinom{14}{4} ="
"1+14+91+364+1001=1471"
c) We have "2^{14}" different bit strings of length 14. 1471 of them contains at most four 1 and 1001 contains four 1. Means bit strings of length 14 that contain at least four 1 is
"2^{14}-1471+1001=15\u00a0914"
d) We should choose 7 place from 14 for 1. And another will fill with 0.
"\\dbinom{14}{7}=\\frac {14!} {7!(14-7)!}=3432"
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