A five digit number is to be formed using each of the digits

Awesome solutions MacFauz and Bunuel, +1 kudos for both. Thanks a lot

Great learning on this one!! thank you very much!

No KUDOS?

Good questions deserves kudos as much as a good solution. +1.

Haha... I second that....

C

Just subtract from the total number of cases.

total number of combinations=5*4*3*2*1=120
When two numbers are together, we will have pack the two numbers together, therefore combinations=4!*2(numbers can be arranged themselves)

The cases where they are not together=total-cases when they are together

120-48=72

Reqd ways = (Total No. of ways - cases when both 1 and 2 are together)

5 numbers can be arranged in 5 places in 5! ways. Now tie 1 and 2 together so effectively there are 4 nos. they can be arranged in 4! ways. 1 and 2 can be arranged within themselves in 2!ways.

Reqd. Answer = 5! - 4!2! = 120 - (24*2) = 72

Answer: C

Another way to ensure that 1 and 2 are not together is to use 'Gaps' method. Basically, when we are given 'n' objects to be arranged so that 'a' out of them should NOT be together - we first arrange the remaining 'n-a' objects, and then arrange these 'a' objects in the gaps created by those 'n-a' objects.

I will try to explain by this question.

We have to arrange 1, 2, 3, 4, 5 so that 1 and 2 are not together. So lets first arrange the remaining numbers.
Remaining 3 numbers can be arranged in 3! = 6 ways.

Now, for each of these 6 ways, the arranged numbers create 4 gaps:
...3...4...5...

one gap before 3, another gap between 3 and 4, another gap between 4 and 5, and the last gap after 5.

So for every arrangement of those 3 numbers, we have 4 gaps - and we can now place the numbers '1' and '2' in any two out of these 4 gaps (and thus they will not touch each other). Number of ways of doing that = 4P2 (arranging 4 objects at 2 places) = 12 ways.

Thus total ways = 3! * 4P2 = 6*12 = 72. Hence C answer

PS - this same can be extended to any such question. So if we have 7 people out of which A,B,C have to be separated (no two of them can sit together), we can first arrange remaining 4 people in 4! ways, and now out of the 5 gaps created (4 people in a line will create 5 gaps), we can place A,B,C in any of the 3 gaps out of 5 in 5P3 ways (60). Thus total ways = 4! * 5P3 = 24 * 60 = 1440.

5!- 4!2!

C

Ans - C.
Total number possible - total number possible with 1 & 2 together =
5! - (4! * 2!) = 120 - 48 = 72