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Using Permutations/ Combinations

The question is: How many $4$ digits no. are divisible by $5$ if no $2$ digits are the same.

The answer is: $952$

Thanks

  • 0
    What have you attempted? Where are you stuck? – 2017-02-26
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    I think it is a permutation question but do not know how to use the formula in this case – 2017-02-26
  • 0
    It is more basic than being a "permutation" question., it is a multiplication principle and addition principle question. – 2017-02-26
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    Consider cases. The last digit is a $0$, or the last digit is a $5$. – 2017-02-26
  • 0
    So it is 9 choices for first no., 8 for 2nd, 7 for 3rd, and 1 for last digit? – 2017-02-26
  • 1
    That is getting closer, but it is different depending on if the last digit is a five or a zero. Do you see why? To word it more accurately, pick the last digit *before* picking the other digits. – 2017-02-26
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    I still do not understand how I am to get to the answer – 2017-02-26
  • 0
    Do you know what the [multiplication principle](https://en.wikipedia.org/wiki/Rule_of_product) is? Do you know what the [addition principle](https://en.wikipedia.org/wiki/Rule_of_sum) is? Add together the number of four digit numbers with no repeated which end in five to the number of four digit numbers with no repeated which end in zero. To count how many end in five, describe them by multiplying number of options available at each step of the following sequence: pick last digit, pick first digit, pick second digit, pick third digit. (*note first digit can't be zero*). Similar for zero – 2017-02-26

4 Answers 4

1

Hint:

A number which is divisible by five ends in either a zero or a five.

A four digit number does not start with zero.

How many four digit numbers with no repeated digits end with a zero? How many end with a five?

0

Number divisible by $5$ has $5$ or $0$ at units place. There are $2$ case.

$1$) $0$ at units place $2$) $5$ at units place

You will also have to ensure for $2$ case that you will not take $0$ at thousands place. So you will have $8$ ways to select digit at thousands place(other than $0$ and $5$),$8$ ways for hundreds place and $7$ ways for the place.... Hope you can proceed of there is any problem tell.

0

Since our number is divisible by $5$, we consider cases where our number ends in a $0$ or a $5$. \begin{array}{ccccc} 1&2& 3 &4&\\ & &&0 \\ & & &5 \end{array}

Consider the first case, where our number ends in $0$. We have three positions that vary freely, with the exception that none can have the same value as another. Note, that since our last digit ends in $0$ we cannot allow our first digit to end in zero. Thus, there are $9 \cdot 8 \cdot 7$ possibilities.

In the other case, where our number ends in $5$, we have three positions to vary freely. However, our first value cannot be $5$ or $0$ (if the first digit were $0$, we would not have a four digit number). Thus, we have $8$ possibilities for our first digit. We have $8$ possibilities for our second digit since we allow for this number to be $0$, but not $5$ or the value of our first digit. For obvious reasons, our third digit has 7 possibilities.

Finally, we arrive at $(9 \cdot 8 \cdot 7) + (8 \cdot 8 \cdot 7) = 952$ possibilities.

0

I know I'm too late for the answer/question but I come up with a different approach and I would want to know if my method is also correct.

Let's start with divisibility rule of 5 where last digit must either be 5 or 10.

So the last digit only have two cases.

There are 4 spots to fill in since it's a 4 digit number.

put 2 on the first spot, No 2 numbers must be the same so we would have 9 more possibilities for the second spot and 8 for the third and 7 on the last

Now we have 2x9x8x7 which equals to 1008

We can't have 0 on the thousands place since it would make your number a 3 digit one. So we must subtract that to the result. This should be 8×7 or 56. Because first spot is already ocuppied by 5 or a 0 and the last is assumed to be taken by 0 again leaving 8 more possibilities x 7 more for the remaining spots.

This gives us 1008-56 or 952 in short.