Aptitude - Permutation and Combination - Discussion

3p3*2C1 = 12.

Since we have four places and the no which is divisible by 4 must have 2 or 6 in the unit place, so 2 and 6 is consider for the unit place, since we are choosing between them so we use combination 2C1.

Now only 3 places are left with three digits 3, 7 and (2 or 6), now here we are arranging them so we use permutation so 3P3.

I need some help on how to come up with those digits?

2, 3, 6, 7, 9 and 5 is must after two digits (because it must divisible by 5). Then we take 5 numbers (2, 3, 4, 7, 9) with 2 groups. Here we applies permutation (because 23 and 32 are different).

5p2 = 5!/(5-2)! = 5!/3! = 4*5 = 20.

Note that ab, ba are two different permutations but they represent the same combination.

How many even four-digit numbers be formed by using the integers 2, 3, 4, 5 without repetition? How many of these numbers will be less than 3000?

There are 6 digits are given ,

As per question one(5) is fixed so 5! = 120.

Three digit number is going to be formed by 3! combination.

So, 120/6 = 20.

Say number 5 is placed in the unit's place, two places, and five numbers are left. Can't it be solved as 5C2?

Thanks @Rahul! good explanation.

Shortcut way to solve this problem :-6C3 => 6!/3!
= 6 * 5 * 4/3 * 2 * 1.
= 20.

Why can't we write 5c2? There are two places and 5 numbers.

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.