Finding amount of numbers.

**Monika1987** · March 9th 2010, 10:49 PM

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.

**Swlabr** · March 9th 2010, 11:54 PM

Originally Posted by Monika1987

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!

**chisigma** · March 10th 2010, 12:28 AM

A quick alternative is the use of 'binomial coefficients'. The number of way to dispose k zeroes in a numer of n digits is ...

$\binom{n}{k} = \frac{n!}{k!\cdot (n-k)!}$ (1)

... so that the number of 8 digits numbers having no more than 3 zeroes is...

$N= \binom{8}{0} + \binom{8}{1} + \binom{8}{2}+ \binom{8}{3}$ (2)

Kind regards

$\chi$ $\sigma$

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