Combinatronics

[adityanarula]

I dont have the OA for this:

There are 20 different scholarships to be given to students at West Month High. How many ways are there for 4 students to win the scholarships?

[spoiler]My solution: 20*19*18*17[/spoiler]

[sreak1089]

Ans should be 20 C 4 = (20 * 19 * 18 * 17 * 16!) / (4! * 16!)
= (20 * 19 * 18 * 17) / 24 = (19 * 18 * 17 * 15)
= 87210 ways

[ashaydesai]

IMO : 20*19*18*17

Its a permutation problem , not combination. Hence its not 20C4.

[amitchell]

I dont have the OA for this:

There are 20 different scholarships to be given to students at West Month High. How many ways are there for 4 students to win the scholarships?

[spoiler]My solution: 20*19*18*17[/spoiler]

Adit et al -

The wording of a question is a little vague - it leaves unclear whether each student wins only exactly one scholarship or can win more than one. Say we go with one scholarship / person. We are pairing distinguishable scholarships with distinguishable recipients, so we have an ordering problem and we use the permutation formula:

[spoiler]n=20 and k=16, so 20!/16!= 20 x 19 x 18 x 17.[/spoiler]

You can also count your way to the solution. There are 20 ways to assign a scholarship to the first person. For each of those ways, there are 19 ways to assign a scholarship to the next person, and so on for 20 x 19 x 18 x 17.

[gospy]

What has been said above is correct. What distinguishes this problem from a combination is the distinct objects to distinct individuals part.

When I took combinatorics, it helped me to make a chart.

For n distinct objects into k boxes with repetition => k^n
For n distinct objects into k boxes without repetition => k!/(k-n)!
For n identical objects into k boxes with repetition => (n+k-1)C(n)
For n identical objects into k boxes without repetition => kCn

In this case there are distinct scholarships going to distinct people without repetition, so we use that case.

If all of the scholarships were identical, it would be last choice.

[ankitns]

I agree that the wording is vague. It could be possible that each student wins multiple scholarships. It could also be possible that a student does not win any scholarships. If the these two conditions can be true, then the answer is 5 to the power of 20.

1. There are 4 students [A, B, C & D]
2. Each scholarship has 5 buckets to go to. Student A, B, C, D or NONE (if nobody wins that scholarship).

Since each scholarship has 5 outcomes...answer would be 5 X 5 X 5....20 times.

hence 5 to the power of 20.

Any thoughts?

I dont have the OA for this:

There are 20 different scholarships to be given to students at West Month High. How many ways are there for 4 students to win the scholarships?

[spoiler]My solution: 20*19*18*17[/spoiler]

Adit et al -

The wording of a question is a little vague - it leaves unclear whether each student wins only exactly one scholarship or can win more than one. Say we go with one scholarship / person. We are pairing distinguishable scholarships with distinguishable recipients, so we have an ordering problem and we use the permutation formula:

[spoiler]n=20 and k=16, so 20!/16!= 20 x 19 x 18 x 17.[/spoiler]

You can also count your way to the solution. There are 20 ways to assign a scholarship to the first person. For each of those ways, there are 19 ways to assign a scholarship to the next person, and so on for 20 x 19 x 18 x 17.