Problem 5

Problem – 5

How many bit strings of length 13 have

a) exactly three 0s?

b) more 0s than 1s?

c) at least ten 1s?

d) at least three 1s?

 

Solution:

a)      Number of possible strings exactly have three 0s = 13C3 = \frac{13\times12\times11}{1\times2\times3} = 286 bit strings

b)      There are 7 possibilities when number of zeros are more than number of ones, so possible bit strings are = 13C13 + 13C12 + 13C11 +13C10 +13C9 +13C8 + 13C7 = 1 +13 +78 +286 +715 +1287 + 1716 = 4096

c)      There are 4 possibilities when at least ten 1s , so possible bit strings are = 13C10 + 13C11 +13C12 +13C13 = 286 + 78 + 13 + 1 = 378

d)     There are two ways to find the possible string  : first one is calculate all possibilities like when three 1s , four 1s and up to Thirteen 1s and add them.

Second ways is calculate total strings and minus the all not possible cases which are only 3.

Here we are following second method = total possible bit strings – not possible strings

= 2^{13} – (13C0 +13C1 + 13C2)

= 8192 – ( 1+13+78)

= 8192 – 92

= 8100 bit strings

 

Leave a Reply