I need help with finding how many 8-digit numbers there are containing no more than three zeros?
For exampel numbers like 10001111.
Thanx in advance.
Can you think of a way of finding out how many numbers have no zeros? one zero? Two zeros? Three zeros?
(Hint: you have 8 places, and you are choosing none to be a zero, then one to be a zero, then two to be a zero, then three to be a zero...)
Finally, subtract these from the total number of numbers to get your result!
A quick alternative is the use of 'binomial coefficients'. The number of way to dispose k zeroes in a numer of n digits is ...
$\displaystyle \binom{n}{k} = \frac{n!}{k!\cdot (n-k)!} $ (1)
... so that the number of 8 digits numbers having no more than 3 zeroes is...
$\displaystyle N= \binom{8}{0} + \binom{8}{1} + \binom{8}{2}+ \binom{8}{3}$ (2)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$