Question

Number of 5 digit numbers which are divisible by 5 and each containing the digit 5, digits being all different is equal to $k\left( 4! \right)$, the value of k is:
A. 84
B. 168
C. 188
D. 208

Answer 1

To do this question, we first need to know the divisibility test of 5 which is given as: the numbers should end in either 5 or 0. Since we have been given that 5 is already one of the digits, we will take two different cases- one which contain 0 and the other which do not. We will solve the case in which 0 isn’t there first by fixing the last position as 5 and then filling the remaining left over positions by the numbers from 1 to 9 except for 5. Then we will solve the second case in which 0 is there, in that, we will make two sub-cases- one in which the number ends with 0 and the other in which the number ends in 5. We will solve them the same way as we solved the first case. Then we will add the outcomes of all these cases and put it equal to k(4!). Then we will solve the equation thus obtained and hence we will get the value of k and thus our required answer.

Complete step by step answer:
We here have been asked to find out 5 digit real numbers which are divisible by 5 and also contain the digit 5 and all the digits are different. For this, we first need to know the divisibility test for 5.
We know that every number which ends in either 0 or 5 is divisible by 5. Now, since we have already been told that the number should contain 5 as one of its digits, we need to find the other 4 digits.
Now, the other 4 may or may not contain 0 as one of its digits. We will first find the number of 5 digit numbers which do not contain 0. This is given as follows:
We know that the number should either end in 0 or 5 to fulfil the required conditions. But since in this case, the numbers do not contain any 0, the number should end in 5 only. Thus, we can write the number as:
$\\_\ \\_\ \\_\ \\_\ \underset{\scriptscriptstyle-}{5}$
Thus, the last position of this number is fixed. Now, the first 4 positions can be filled by any number between 1 and 9 except for 5 but they all should be different. Thus, we need to select and arrange 3 numbers out of 9-1=8 numbers.
We know that the number of ways to first select and then arrange ‘r’ things out of ‘n’ things are given as $^{n}{{P}_{r}}$ which is equal to $\dfrac{n!}{\left( n-r \right)!}$ .
Here, we can see that:
$\begin{aligned} & n=8 \\\ & r=4 \\\ \end{aligned}$
Thus, the first 4 positions can be filled in the following number of ways:
$\begin{aligned} & ^{8}{{P}_{4}} \\\ & \Rightarrow \dfrac{8!}{\left( 8-4 \right)!} \\\ & \Rightarrow \dfrac{8!}{4!} \\\ & \Rightarrow \dfrac{8\times 7\times 6\times 5\times 4!}{4!} \\\ & \Rightarrow 8\times 7\times 6\times 5 \\\ \end{aligned}$
Hence, the number of 5 digit numbers which are divisible by 5 which do not contain 0 are given by $8\times 7\times 6\times 5$.
Now for the numbers which do contain 0, the last position can be filled by either 0 or by 5.
Case-1: last position is filled by 5
This number is shown as follows:
$\\_\ \\_\ \\_\ \\_\ \underset{\scriptscriptstyle-}{5}$
Now, among the other 4 positions, 0 can occupy any but the first position. Hence, it has 3 ways in which it can be positioned. Now, the remaining three positions can be filled by 8 digits in $^{8}{{P}_{3}}$ ways. Thus, the total 5 digit numbers which end in 5 and contain 0 and are divisible by 5 are given as:
$\begin{aligned} & 3{{\times }^{8}}{{P}_{3}} \\\ & \Rightarrow 3\times \dfrac{8!}{\left( 8-3 \right)!} \\\ & \Rightarrow 3\times \dfrac{8\times 7\times 6\times 5!}{5!} \\\ & \Rightarrow 3\times 8\times 7\times 6 \\\ \end{aligned}$
Case-2: last position is filled by 0
This number is shown as follows:
$\\_\ \\_\ \\_\ \\_\ \underset{\scriptscriptstyle-}{0}$
Now, among the remaining 4 digits, one should be 5 and the rest can be anything. 5 can take any of the 4 positions and the remaining 3 positions can be filled in $^{8}{{P}_{3}}$ ways.
Thus, the number of 5 digit numbers which contain 5 and end in 0 are given as:
$\begin{aligned} & 4{{\times }^{8}}{{P}_{3}} \\\ & \Rightarrow 4\times \dfrac{8!}{\left( 8-3 \right)!} \\\ & \Rightarrow 4\times \dfrac{8\times 7\times 6\times 5!}{5!} \\\ & \Rightarrow 4\times 8\times 7\times 6 \\\ \end{aligned}$
From all the above calculations, we can see that the number of 5 digit numbers which contain 5 and are divisible by 5 are given as:
$\left( 8\times 7\times 6\times 5 \right)+\left( 8\times 7\times 6\times 3 \right)+\left( 8\times 7\times 6\times 4 \right)$
Now taking $8\times 7\times 6$ common we get:

& \left( 8\times 7\times 6\times 5 \right)+\left( 8\times 7\times 6\times 3 \right)+\left( 8\times 7\times 6\times 4 \right) \\\ & \Rightarrow \left( 8\times 7\times 6 \right)\left( 5+4+3 \right) \\\ & \Rightarrow 12\left( 8\times 7\times 6 \right) \\\ \end{aligned}$$ Now, since we have been given this equal to k(4!), we can say that: $12\left( 8\times 7\times 6 \right)=k\left( 4! \right)$ Solving this we get: $\begin{aligned} & 12\left( 8\times 7\times 6 \right)=k\left( 4\times 3\times 2\times 1 \right) \\\ & \Rightarrow 24k=12\left( 8\times 7\times 6 \right) \\\ & \therefore k=168 \\\ \end{aligned}$ Hence, the value of k is 168. **So, the correct answer is “Option B”.** **Note:** The places at which we have used the formula $^{n}{{P}_{r}}$, we can also use the combination of two different formulas which result the same. We can do this by first selecting and then arranging. We know that the number of ways for selecting ‘r’ objects out of n distinct objects is given as $^{n}{{C}_{r}}$ and we also know that r distinct objects can be arranged in r! ways. Thus, the number of ways of first selecting and then arranging r objects out of n is given as: $\begin{aligned} & ^{n}{{C}_{r}}\times r! \\\ & \Rightarrow \dfrac{n!}{r!\left( n-r \right)!}\times r! \\\ & \Rightarrow \dfrac{n!}{\left( n-r \right)!} \\\ & {{\therefore }^{n}}{{P}_{r}} \\\ \end{aligned}$