Count number of binary strings

For a string of binary digits of length 11 11 bits, how many strings are there with at least Four 0 s 0's and at least Four 1 s 1's .


The answer is 1584.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

David Vreken
Dec 14, 2018

A string of binary digits of length 11 11 bits with at least four 0 0 's and four 1 1 's has three bits leftover that could be either all three 0 0 's, two 0 0 's and one 1 1 , two 1 1 's and one 0 0 , or all three 1 1 's. These have the following total number of different anagrams:

0000 1111 000 11 ! 7 ! 4 ! = 330 0000 \text{ } 1111 \text{ } 000 \rightarrow \frac{11!}{7!4!} = 330

0000 1111 001 11 ! 6 ! 5 ! = 462 0000 \text{ } 1111 \text{ } 001 \rightarrow \frac{11!}{6!5!} = 462

0000 1111 011 11 ! 6 ! 5 ! = 462 0000 \text{ } 1111 \text{ } 011 \rightarrow \frac{11!}{6!5!} = 462

0000 1111 111 11 ! 7 ! 4 ! = 330 0000 \text{ } 1111 \text{ } 111 \rightarrow \frac{11!}{7!4!} = 330

for a total of 2 ( 330 + 462 ) = 1584 2(330 + 462) = \boxed{1584} different strings.

So, leading digits are allowed, right?

Henry U - 2 years, 5 months ago

Log in to reply

Yes, typically when using binary notation of n bits it is conventional to include cases with leading zeros.

David Vreken - 2 years, 5 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...