# For those of you who know combinatorics, can you solve this problem?

Mariah recently gave a combinatorics problem that she was working on, and it reminded me of this problem. The twist needed to solve it is not usually taught in an intro class, but you should be able to figure it out.

I am sure you have all seen the type of test problem where you are given two columns of entries and you are asked to match the entries in the first column with those in the second. Suppose that you are given 6 rock bands in one column and the names of 6 songs in the second. Each band corresponds to one song. If you chose at random, what are the chances of getting at least 3 correct? You can’t use a band or song more than once.

Observing members: 0 Composing members: 0

I would think P3c = 1/6 * 1/7 * 2/17.

WyCnet (184)

My 1st answer was incorrect, here is the adjusted value: P3c = 1/6 * 1/5 * ¼ = 1/120. The reason why my 1st answer was incorrect was because correct pairs are removed from the population for sucessive tries.

WyCnet (184)

Thanks for answering. I did not think anyone was going to.

Here is how to solve this.

To find the number of ways of getting exactly 3 right, we start with C(6,3). That is the number of ways of choosing 3 questions that are answered correctly. However, to get exactly 3 right, none of the others can be correct. In each case we can designate the remaining questions as q1, q2 and q3 and the remaining answers as a1,a2,and a3. We want to know in how many ways we can match a1, a2 and a3 to q1,q2 and q3 so that none of the answers is correct. This is the same as asking how many ways can we arrange 1, 2 and 3 so that none of the numbers is in the same spot.This type of permutation is called a derangement. For large numbers, it is a bit tricky to count them, but for 3 numbers, it is easy to see that there are just 2 – (2,3,1) and (3, 1, 2). That gives us C(6,3)*2.

Similarly, the number of ways of getting exactly 4 right is C(4,2) times the number of derangements of 2 numbers. There is only one such permutation. For example, for 1 and 2 it would be (2,1) That gives C(4,2) * 2

There are no ways of getting exactly 5 right, because if you choose 5 right, the last one will automatically be right also. In terms of derangements, what we are saying is that there are no derangements of one number.

Finally, there is exactly one way of getting all 6 right.

Adding these together gives (C6,3)* 2 + C(4,2) + 1 = 56
The probability then is 56/6! = 56/720 = 7.8% to one decimal place.

I should have said C(4,2)*1 instead of C(4,2)*2

I started believing my second answer was incomplete because the first pick could have been wrong, which meant I had to multiply twice more for two incorrect picks for the 2nd answer to become correct for only 3 picks. This meant using ⅔ and ½, which gives 1 over 360 for only three in the order C-C-C-W-W-W, where C is correct and W is wrong.

If the 1st answer is Wrong, which has a probability of 5/6 then there can only be 3 correct, so to have 4 or more correct, the chooser has to guess right from the get go.

W-C-C-C-W-W is 5/6 * 3/16 * 2/9 * ¼
C-C-C-W-W-W is 1/6 * 1/5 * ¼ * ⅔ * ½
which are different!

WyCnet (184)

Got the last one wrong again. OMG
W-C-C-C-W-W is 5/6 * 4/25 * 3/16 * 2/9 * ¾
C-C-C-W-W-W is 1/6 * 1/5 * ¼ * ⅔ * ¾
which are the same. I should have stuck with the formula.

WyCnet (184)