# binomial problem!

• Jan 13th 2012, 03:56 AM
lawochekel
binomial problem!
5 digits numbers is formed by using the digits 0,1,2,3,4 and 5 without repetition. find the probability that the number is divisible by 6?

pls help out here, confuse!
• Jan 13th 2012, 05:23 AM
Soroban
Re: binomial problem!
Hello, lawochekel!

I'll explain it using baby-steps.

Quote:

Five-digit numbers are formed by using the digits {0,1,2,3,4,5} without repetition.
Find the probability that the number is divisible by 6.

There are $6!= 720$ possible five-digit numbers.

To be divisible by 6, the number must be:
. . (1) divisible by 3 ... the sum of its digits must be divisible by 3.
. . (2) even ... the last digit must be even.

There are two sets of digits whose sums are divisible by 3:
. . $\{0,1,2,4,5\}\:\text{ and }\,\{1,2,3,4,5\}$

Consider $\{0,1,2,4,5\}$
The last digit must be even: $\{0,2,4\}$, 3 choices.
The other four digits can be arranged in $4!$ ways.
Hence, there are:. $3\cdot4! \,=\,72$ numbers divisible by 6.

Consider $\{1,2,3,4,5\}$
The last digit must be even: $\{2,4\}$, 2 choices.
The other four digits can be arranged in $4!$ ways.
Hence, there are:. $2\cdot4! \,=\,48$ more numbers divisible by 6.

Hence, there are:. $72 + 48 \,=\,120$ numbers which are divisible by 6.

Therefore:. $P(\text{div by 6}) \:=\:\frac{120}{720} \:=\:\frac{1}{6}$

• Jan 13th 2012, 07:07 AM
Plato
Re: binomial problem!
Quote:

Originally Posted by Soroban
There are $6!= 720$ possible five-digit numbers.

Are you counting $05421$ as a five-digit number?