I came across the question in the picture on the net. I couldn't figure
out the solution given. But what I got is total number of such $5$
digit numbers using the given digits will be $5.5!$. Half of them will
be divisible by $2$. And $1/3$ of them will be divisible by $3$. (Not sure though) What's next? I couldn't decipher what's written after that…
Please try to use the permutations and combinations approach if possible:
A $5$-digit number is formed by using the digits $0$, $1$, $2$, $3$,
$4$ & $5$ without repetition. The probability that the number is
divisible by $6$ is:(A) $8\%$
(B) $17\%$
(C) $18\%$
(D) $36\%$
Ans: (C)
Hint: Number should be divisible by $2$ and $3$.
$n(S) = 5 \cdot 5!$; $n(A)$: reject '$0$' $= 2 \cdot 4!$
reject '$3$' $= 4! + 2 \cdot 3 \cdot 3!$
Total $n(A) = 3 \cdot 4! + 6 \cdot 3! = 18 \cdot 3!$
$\therefore p = \frac{18 \cdot 3!}{5 \cdot 5!} = 18\%$
Best Answer
We try to count how many five-digit numbers formed with digits $0,1,2,3,4,5$ are divisible by six.
To begin, we note three important facts:
So, putting this information together, we start counting.
Case 1: $0$ is not used
Case 2a: $3$ is not used and $0$ is at the end
Case 2b: $3$ is not used and $0$ is not at the end
There are then a total of $2\cdot 4!+4!+2\cdot 3\cdot 3! = 108$ possible five-digit numbers satisfying the conditions.
The total number of five-digit numbers that can be created with those available digits will be $5\cdot 5!=600$. (Arrange all six digits in a row, the final digit is unused. Remove the "bad" arrangements where zero is the leading digit for $6!-5! = 5\cdot 5!$)
The probability that a five-digit number formed from those digits is divisible by six is then $\frac{108}{600}=0.18$