Aptitude - Permutation and Combination - Discussion

The greatest 4 digit number which is only divisible by 5.

n!/(n-r) !.r!, is the best way to solve the problem.

The 3 digit no should be divisible by 5 so it must the 5 is at the unit place then the 3 digit no will be ---- xy5.

Now the two no. (x and y ) are taking from five ( 2,3,6,7,9).

So 5p2 = 5*4 = 20.

Friends,
How many 4 lettered words divisible by 4 can be formed fro 0 1 2 3 4 5?

And if repetition of digits is not allowed?

The correct answer is 60 ways. The logic behind this is, we know that when a number is divisible by 3 the sum of the digit should be divisible by 3. In the given number, only 2,3,6,7,9 are divisible by 3 while adding. 2+3+6+7+9=27,
so 5*4*3=60 ways.

I hope that it is right. If it is wrong please help me.

5 has to come in the last digit. So its fixed. Now we are concerned only about the first two digits. We can choose from remaining five no's in 5C2 ways. And this can be arranged in 2! ways. So ans is 5C2*2! = 20.

Since it says that we have to look for numbers that are divisible by 5 among 2,3,5,6,7,9 so the unit digit has to be 5.
__ __ _5_
Since we don't know which numbers are going to be on the first and second (hundredth and tenth) we have to exclude 5 and think about all the cases that can be above.

2,3,6,7,9 are left except 5. So 5 numbers can go into the first digit which is hundredth, and since we used 2 numbers so far, we subtract 2 from the total number of the numbers on the top.

6-2 = 4 (left).

So in the second digit, only 4 numbers can go in.
Hence, 5*4*1 = 20 (answer)

@Raj.

Given Digits: 2,3,5,6,7,9

If the digits are repeated then

Let 3 digit no. be XYZ.

Divisible by 5 must have 5 at Z place or unit place i:e XY5. that is only one way to fill Z place.

Now Y place can be filled by any of 6 digits i:e 2,3,5,6,7,9
in 6C1 = 6 ways.

Also X place can be filled by any of 6 digits i:e 2,3,5,6,7,9
in 6C1 = 6 ways.

So total ways = 6*6*1 = 36 ways.

I think its clear to you know. :).