Hi, there. I'm happy to help with this.

You ask for a solution with the "permutation formula only" --- hmm, it's slightly harder to solve that way. I'll show what I consider two easier solutions, and then a harder solution with permutations.

Solution #1:

This one is easier, and involves the Fundamental Principle of Counting.

Choices for the tens digit = 5

Choices left for the units digit = 4

Total number of choices = 5 * 4 = 20.

Solution #2:

It's relatively easy simply to enumerate every possibility:

12, 13, 14, 15,
21, 23, 24, 25,
31, 32, 34, 35,
41, 42, 43, 45,
51, 52, 53, 54 --> 20 possibilities.

Solution #3:

As requested, a solution with permutations.

nPr = the number of permutations of r elements taken from a set of n

nPr = (n!)/((n-r)!), where n! is the factorial symbol.

Here, we are selected 2 items form a pool of 5, and order matters ---- 23 is a different number from 32. Therefore, it is a permutation, and the total number of permutations is:

5P2 = (5!)/(3!) = (5*4*3*2*1)/(3*2*1) = 5*4 = 20 possibilities

So, three solutions, each giving the same answer. I hope that helps.

Does it all make sense? Please do not hesitate to ask if you have any questions on what I've said.

Mike

Dear Abhinav Khanna,

That's a great question.

The reason we don't have to subtract 5 is because of the way permutations are designed. When we take a permutation of k elements from a pool of n, implicit in that process is that (a) the objects distinct from one another, and (b) we are selecting them without replacement. If Q is an element of the pool of n, then Q will either be selected or not selected, but if it is selected, there is no possibility of its being selected twice.

If you think about it, that's where the whole factorial idea comes from for permutations. For example, consider 4P4, the permutations of a set of four distinct objects --- 4P4 = 4! = 4*3*2*1 = 24. The reason is: for the first position, you have four choices; then for the second position, you have three choices --- the three remaining after the first choice; for the third position, you have two choices, the two remaining after the first two positions are chosen; then, only one choice remains for the last position. That is all assuming --- no object is going to be chosen more than once: when an item is chosen, it is "used up", no longer available. This means: no repeats.

If we could repeat as much as we wanted, then there would be four choices available for each of the four places, and the total number would be 4^4 = 256. That's something quite different from a permutation.

If you want to allow for the possibility of repeat items --- for example, digits where double digits are allowed, then permutations are not the appropriate tool to use. You would either have to (a) calculate the number of permutations, then add in the doubles, or (b) just stick with the Fundamental Counting Principle (which is always a port in a storm).

Here's a free video about the FCP.

https://gmat.magoosh.com/lessons/335-fun ... -principle

Again, that was a very perceptive question. Does all that I said make sense to you? Please do not hesitate to ask if you have any further questions.

Mike