1. ## 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!

2. ## Re: binomial problem!

Hello, lawochekel!

I'll explain it using baby-steps.

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 $\displaystyle 6!= 720$ 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:
. . $\displaystyle \{0,1,2,4,5\}\:\text{ and }\,\{1,2,3,4,5\}$

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

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

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

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

3. ## Re: binomial problem!

Originally Posted by Soroban
There are $\displaystyle 6!= 720$ possible five-digit numbers.
Are you counting $\displaystyle 05421$ as a five-digit number?