# A five digit number is formed by the digits \[1,2,3,4,5\] with no digit being repeated. The probability that the number is divisible by $4$, is

$A$. $\dfrac{1}{5}$

$B$. $\dfrac{2}{5}$

$C$. $\dfrac{3}{5}$

$D$. $\dfrac{4}{5}$

Answer

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Hint: - The number of possibilities in each of these cases is $3!$ as the digits at one's and ten's places are fixed and the rest $3$ digits can be chosen in $3!$ ways. Because we know that rearranging the n numbers on n places is $n!$.

We know that the number is divisible by $4$ only when it ends with: when the last two digits are divisible by $4$.

So by the given numbers we can form number which is divisible by $4$ only when it ends with:

\[12,\;,\;24,,\;32,\;\;52\]

The number of possibilities in each of these cases is $3!$ as the digits at one's and ten's places are fixed and the rest $3$ digits can be chosen in $3!$ ways. Because we know that rearranging of n numbers on n places is $n!$.

Thus total possibilities for digit to be divisible by \[4 = 4 \times 3!\] (because we made four conditions above for divisibility by 4).

The total number we can form by these five numbers are: - $5!$ , (arranging 5 numbers in 5 places).

Let $E = {\text{the number is divisible by}}\;4$

$P(E) = \frac{{{\text{Numbers divisible by}}4}}{{{\text{Total numbers}}}}$

$ = \frac{{4 \times 3!}}{{5!}} = \frac{{4 \times 3 \times 2 \times 1}}{{5 \times 4 \times 3 \times 2 \times 1}} = \frac{1}{5}$

Hence, option A is the correct answer.

Note: - Whenever we face such a type of question first find out the favorable outcome. Here we can find out the number which is divisible by four by the divisibility rule by fixing the last two numbers which are divisible by four and rearrange the first three numbers to find a favorable outcome.

We know that the number is divisible by $4$ only when it ends with: when the last two digits are divisible by $4$.

So by the given numbers we can form number which is divisible by $4$ only when it ends with:

\[12,\;,\;24,,\;32,\;\;52\]

The number of possibilities in each of these cases is $3!$ as the digits at one's and ten's places are fixed and the rest $3$ digits can be chosen in $3!$ ways. Because we know that rearranging of n numbers on n places is $n!$.

Thus total possibilities for digit to be divisible by \[4 = 4 \times 3!\] (because we made four conditions above for divisibility by 4).

The total number we can form by these five numbers are: - $5!$ , (arranging 5 numbers in 5 places).

Let $E = {\text{the number is divisible by}}\;4$

$P(E) = \frac{{{\text{Numbers divisible by}}4}}{{{\text{Total numbers}}}}$

$ = \frac{{4 \times 3!}}{{5!}} = \frac{{4 \times 3 \times 2 \times 1}}{{5 \times 4 \times 3 \times 2 \times 1}} = \frac{1}{5}$

Hence, option A is the correct answer.

Note: - Whenever we face such a type of question first find out the favorable outcome. Here we can find out the number which is divisible by four by the divisibility rule by fixing the last two numbers which are divisible by four and rearrange the first three numbers to find a favorable outcome.

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