# Math Help - Binary combination/permutation help

1. ## Binary combination/permutation help

i have some questions involving permutation/combinations and need your help.

here it goes:

im tring to find out how many different arrangements of 4 (1's) can be put in 40 (0's)

ex:
: 0000000000000000000000000000000000000000 empty
1 : 1111000000000000000000000000000000000000
2 : 0000000000000000000000000000000000001111
3 : 0000010001000000000000001000000000000100
etc....

how do i calculate that? O.o

also what is the chance of finding specific one of those in 1000000 random attempts.

any idea's?

2. ## Re: Binary combination/permutation help

Your task can be rephrased to distributing 35 objects into 5 buckets. ${5+35-1} \choose {5-1}$

3. ## Re: Binary combination/permutation help

hmmmm thanks for reply but im getting the feeling thats not it ^^

i think i found out how to do it, but can someone explain to me how to calculate it exactly?

Math Forum - Ask Dr. Math

4. ## Re: Binary combination/permutation help

Yours is the "Empty Urns are Allowed" case.

5. ## Re: Binary combination/permutation help

Hellom, zipzaaaap6656!

(a) How many different arrangements are there for four 1's and forty 0's?

We have a 44-digit number.
The four 1's can be placed in . ${44\choose4} \,=\,135,\!751$ ways.
The forty 0's are placed in the remaining positions.

Therefore, there are 135,751 possible arrangements.

(b) What is the probability of finding a specific number in 1,000,000 random attempts.?

I'll assume that the selection is made with replacement.

$P(\text{not-number}) \:=\:\frac{135,\!750}{135,\!751}$

$P(\text{not-number in 1,000,000 tries}) \:=\:\left(\frac{135,\!750}{135,\!751}\right)^{1,0 00,000} \;=\;0.000\: 632\, 104$

$P(\text{number at least once in 1,000,000 tries})$
. . . . . . $\;=\;1 - 0.000\:632\,104 \;=\;0.999\,367\,896$