How many bit strings of length 10 have:?

Question

exactly three 0's
more 0's than 1's
at least seven 1's
and at least three 1's?

Answer 1

The number of bit strings of length 10 with n 0's (or n 1's in fact):
is C(10,n) , where C(a,b) = a! / [(a-b)!b!] is the combinitorial function.

So the first answer is C(10,3) = 120

In the second, there could be 6,7,8,9, or 10 zeros, so:
C(10,6) + C(10,7) + C(10,8) + C(10,9) + C(10,10)
= 210 + 120 + 45 + 10 + 1 = 386

In the third, C(10,7) + C(10,8) + C(10,9) + C(10,10)
120 + 45 + 10 + 1 = 176

In the fourth,
C(10,3) + C(10,4) + C(10,5) + C(10,6) + C(10,7) + C(10,8) + C(10,9) + C(10,10)
= 2^10 - C(10,2) - C(10,1) - C(10,0)
= 1024 - 45 - 10 - 1
= 968

Answer 2

there are ten position important places in the bit string. each holds a 0 or 1. All the 0's are interchangeable. All the 1s are interchangeable.

3 0's = 10! / 7! 3!

there are 10! / 5! 5! numbers with the same nuber of 0's as 1's. there are 10! strings altogether. So 10! - 10! /5! 5! have either more 0's than 1s or more 1's than 0;s. and there must be equal number of each ... so ( 10! - 10!/5!5!)/2

10!/7!3! + 10!/8!2!

10! - ( 10!/10!0! + 10!/9!1! + 10!/8!2!)

Answer 3

This is a job counting.

use combinations and permutations.

for the first 10C3 or 10!/((7!)(3!) which is 120

More 0s than 1s that would be the same as exactly 6 0s + exactly 7 0s plus exactly 8 0s plus exactly 9 0s plus exactly 10 0s

same drill

10C6 + 10C7 + 10C8 + 10C9 + 10C10

10!/((6!)(4!) + 10!/((7!)(3!)) + 10!/((8!)(2!)) + 10!/((9!)(1!))+10!/((10!)(0!))=

210+120+45+10+1=386

At least 7 means 7 or more

10C7+10C8+10C9+10C10=120+45+10+1=176

at least 3 ones = it's easier to do this with 0s so we can restate the problem to say at most 7 0s and then subtract this from the total of all combinations which is 1024

1024-(10C7+10C8+10C9+10C10)=1024-176=848