Aptitude - Permutation and Combination - Discussion

Here, the answer is 20.

By using the fundamental principle of counting ie, if one activity can be done in 'm' num of different ways and the second activity can be done in 'n' number of different ways and these two activities can be done in MxN different ways.

For example, how mane 3 digit number can be formed By using the numbers 2, 3, 5, 8.without repeating.
Soln:-
Let the numbers are 2, 3, 5, 8 and u all know 3 digit number means it must have unit tens and hundred places.
So, now you just Make three column like H T U.
2 3 4 ie in unit place you can fill any of the 4 numbers means 4 different ways and remaining 2 place that is tens and hundred and in the tens place, you can fill in 3 different ways because repetition not allowed.
If once you used the digit you cannot use again. And next 100 places you can fill in 2 different ways.

Therefore total no of ways = 2 x 3 x 4.
= 24ways.
I Hope you understood.

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)

It explained in a bit confusing way...

This calculation can be explained in a simpler way....

As the 3 digit number must be a multiple of 5, it must have 5 in its units place

Ex: xy5. X and Y can be any of the remaining 5 digits(2,3,6,7,9).

These remaining 5 digits can have 5p2 permutations ( P- because the arrangement is important).

That gives 20 possible arrangements of the 5 digits.

The number of possible permutations of the digit '5' is 5p5 = 1.

Therefore 5p2 and 5p5 ( 20 * 1) gives 20.

Here's the answer!

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.

Well its very simple to think as my frnds says that we shd find 3 values and three place in three digit it means unit place tens place and hundreds place as they told the no. is divisible by 5 so the no. at unit place is 5

So now we are having only 2 place remaning i:e tens and hundreds so while using factorial permutation npr n!/(n-r)! were n + number we have i;e 5 r= the sum of places i;e 2 so 5!/(5-2)!=5!/3!=5*4*3*2*1/3*2*1=5*4 =20

@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. :).

If the three digits must be divisible by 5 and no digit is to be repeated, therefore, it's safe to reserve a spot for 5 and then have 2,3,6,7,9 combine the first two digits (XY5),
Permutation of the first two digits gives 5C2
5!/3!

23,26,27,29
32,36,37,39
62,63,67,69
72,73,76,79
92,93,96,97

These outcomes combine with 5 each. 5P3*1 resulting in 20 outcomes.

Remember, no repetition, so there

The answer can also be solved in another way:
To be divisible by 5, 5 will always be in units place. That leaves two numbers to be chosen for tens and hundreds place out of the remaining 5. Order matters here since 675 is different from 765. So, we will be choosing permutations which states that without repetition the number of permutations are n!/(n-r)! which is 5!/(5-2)! = 5!/3! = 5 x 4 = 20

@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.

The 3-digit no. may be" XY5".

At 100th position 5 is fix because it divisible by 5.

Hence 5 occur only once.

Next 10th place Y it may be 2, 3, 6, 7, 9.

Any one of the no. Can occur hence it is 5 i.e. any one can occur.

Next 1st position X can be filled with remaining 4 no. Because of no repetition i.e the chance is 4.

The 3-digits is -1*5*4 = 20.