Problem Solving

Judges will select 5 finalists from the 7 contestants entered in a singing competition. The judges will then rank the contestants and award prizes to the 3 highest ranked contestants: a blue ribbon for first place, a red ribbon for second place, and a yellow ribbon for third place. How many different arrangements of prize-winners are possible?

10
21
210
420
1,260

The integers from 1 to 100 inclusive are each written on a single slip of paper and dropped into a jar. If one slip of paper is removed at random, what is the probability that the number on it is neither even nor a multiple of 3?

83%
67%
50%
33%
17%

The integers from 1 to 100 inclusive are each written on a single slip of paper and dropped into a jar. If one slip of paper is removed at random, what is the probability that the number on it is neither even nor a multiple of 3?

83%
67%
50%
33%
17%

binit wrote:Hi everybody,

The judges Qn, though an easy Qn, was actually very interesting.
What I did right after seeing the problem: N = 7C5*5C3 = 210
Then I thought why should 5 matter at all, I did: N = 7C3*3! = 210
After some moments I tried: N = 7P3 = 210 (since every constant gets a different colored ribbon, order matters, assigning ribbon is just like numbering)
With FCP, as Mich said, N = 7*6*5 = 210 (easiest this way)
I would like to know whether there is any generalization, towards which we are approaching.. Experts' comments needed.

~Binit.

Your first approach worked because for these particular integers (n choose k) * (k choose r) = n permute (n - r). (Try the problem with n = 8 people instead of n = 7, however.)

Your second and third approaches gave the same answer because for any valid n and k (n choose k) * k! = (n permute k).

Your fourth answer worked because it IS the permutations formula, just written in a format that happens work nicely when n and (n - k) are very close. (That is, 7 permute 3 is perhaps better written as 7*6*5, while 70 permute 30 probably should be left as is.)

GMATGuruNY wrote:The second problem should read as follows:

The integers from 1 to 100 inclusive are each written on a single slip of paper and dropped into a jar. If one slip of paper is removed at random, what is the probability that the number on it is neither even nor a multiple of 3?

83%
67%
50%
33%
17%

One approach:
Total integers = (evens) + (multiples of 3) - (integers that are BOTH even AND divisible by 3) + (neither).

Total integers = 100.

Since half of the integers between 1 and 100 inclusive are even, the number of EVEN integers = 50.

Since 100/3 = 33.33, the number of multiples of 3 between 1 and 100 inclusive = 33.

An integer that is BOTH even AND divisible by 3 is a MULTIPLE OF 6.
Since 100/6 = 16.66, the number of multiples of 6 between 1 and 100 inclusive = 16.

Plugging these values into the equation above, we get:
100 = 50 + 33 - 16 + N
N = 33.

Thus:
P(neither even nor a multiple of 3) = 33/100 = 33%.

The correct answer is D.

Amrabdelnaby wrote:Could you please tell me whats wrong with my thinking process?

P(not even or not not divisible by 3) = p(not even) = p(not even) + p (not divisble by 3) - P ( not even & not divisible by 3)