Aptitude - Permutation and Combination - Discussion

Hi..
my ques is that why did we not do 5! x 4! x 1!
as per which it should be 5*4*3*2*4*3*2*1
and not 5*4*1.........

and secondly why did we go for permutation over combination here??

Hey my doubt is not related to this qus. But a general one. What is the diff between permutation and combination in terms of logic, not formulae.

Can also be solved as :---

Units place is fixed so only one method for it.

Now for tens and hundreds place....we can use Permutations as here ordering of digits matters.
As 5 is already reserved at units place so we are left with two choices..
So it becomes :

5P2 * 1

5!
= ------ * 1 = 5x4 = 20 Ans..
(5-2)!

(mark that none of the digits are repeated)
We must use 5 in units(bcz to b divisible wid 5 only/)
then we have 4 other nos. to use in tens then
again to avoid repetation of no.s we should use 3 nos in the hundreds place.
so according 2 me ans is 1*4*3=12

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

Thanks karthika and ashu.

How many five digit positive integers that are divisible by 3 can be formed using the digits 0, 1, 2, 3, 4 and 5, without any of the digits getting repeated in a number?

Can any one explain what is the procedure if the digits are repeated for the above problem?

As the requried condition is XY5.

Since 5 is fixe so we have to select two nos (X & Y) out of remaining five nos i.e,
5C2 & Two nos X&Y arrange in 2! Ways.
=> Total no. Way= ⁵C₂* 2!

= (10)*2 = 20

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!