Aptitude - Permutation and Combination - Discussion

_ _ 5.

There is 2 chance only,

So we use the permutation formula for it,

NPr=n!/ (n-r) !
5P2 = 5!/ (5-2) !
= 5 * 4 * 3 * 2 * 1 /3 * 2 * 1.
= 5 * 4.
= 20.

It is 5C2 * 2! = 20.

H T O
_ _ 5.
So two chance only.
Therefore ans 5C2 = 5 x 4 = 20 simple.

We have to form 3-digit numbers which are divisible by 5,
In that case in one's place, there should be 5, there is only one way to place 5 at one's place
So, we have already selected 5 among 2,3,5,6,7,9 then we have only 2,3,6,7,9
there are 5 ways to place a number among 2,3,6,7,9 numbers at ten's place.
So we have to place any number among 2,3,6,7,9 at ten's place, then the remaining numbers we have only 4,
So we have four ways to place any of the numbers in hundred's place.
Therefore the total ways 1*5*4=20.

@Dinesh.

It is 5P2 = 5*4 =20.
If it is 5C2 = 5*4/2*1 =10.

@Sarvani.

So we have 3 digits, so if the 1s place is 5 we can say the number can be divisible by 5.

Now the number of ways we can fill the 1s place is ( 1 ) _ _ 1W . now we have 2,3,6,7 and 9 so we can fill 100th place in 5 ways.

So we have 5W _ 1W. Now we are left with 10th place and we have only 4 distinct numbers so we can write it as 5W 4W 1W which is nothing but 20.

Thanks @Sujit.

Please explain it clearly. I didn't understand.

@ Vinay.

If you observe the digits given in question there is no '0'. So at units place, we won't consider '0'. And once that is 5. And each of them has 4 ways.
So, 5 * 4 = 20 ways.

Can anyone say is 20 the right answer?

Because the question asked is number divisible by 5 and also not repeated?