Colored bottles

Let the bottle caps be A, B, and C

The possible ways for the caps to NOT be in the exact order of A,B and C are:

B, C, A

C, A, B

= 2 total ways.

Any one of the bottle caps can only be placed on 2 out of 3 bottles so that the colors do not match. Once one bottle cap is placed, there is only one way for the others to be arranged so that they are not on a bottle of their color.

Using derangement theorem No. of permutations = n!(1 - 1/1! + 1/2! - ..... (-1)^n 1/n!) = 3!(1-1/1! + 1/2! - 1/3!) = 2

If we want to find the # of ways that none is correct, there fore... That is... # of permutation on 2 caps is correct / the permutation of 1 cap is correct...

So, n!/(n-r)! So, if you want to find the permutation of 2 caps is correct, 3!/(3-2)! = 3!/1! = 6

And the permutation of 1 cap is correct, n!/(n-r)! 3!/(3-1)! 3!/2! = 3...

So, in sum, 6/3 = 2....

So, the answer is 2

We can use PIE (Principe of Inclusion-Exclusion )

The number of ways can these caps be put on the bottles such that none of the caps are on the correct bottles= All the possible permutations- permutation where at least one cap is in its correct bottles.

All the possible permutations: 3!

Permutation where at least one cap is in its correct bottles: $\binom{3}{1} * 2! -\binom{3}{2}* 1! +\binom{3}{3}*0!$

Therefore the answer would be : $3!-(\binom{3}{1} * 2! -\binom{3}{2}* 1! +\binom{3}{3}*0!) = 6 - (3*2 -3+1) = 2$

• The first bottle's cap can be placed in 2 ways ( excluding its own)
• The second bottle's cap can be placed in 1 way ( excluding its own)
• The third bottle's cap can be placed in 1 way ( whatever is left)

So 2 x 1 x 1 =2

This is a problem about derangement. I've written a pretty long note on this on my dead blog (link below, not advertising because I don't even use that blog anymore). Check it out if you want to know about the generalization of this problem. I wrote it mainly for myself so please don't be too critical.

It talks about a couple of approaches on arriving at a generalization that involves n bottles and n bottle caps. Of course, none of that is necessary for this problem because it can easily be solved by simple listing.

Link to the blogpost: http://dailymathchallenge.blogspot.sg/2013/11/derangement.html

3d1=0 , 3d2=1 , 3d3=2

n! are the number of permutations (ways to order the caps) That would be 3! = 6 ABC, ACB, BAC, BCA, CAB, CBA

From those we have to subtract the ones where some caps are on the right bottle: 3 Caps are right: The only way this can happen is with the original ordering (ABC) 2 Caps are right: This is impossible as it can't be achieved by swapping caps 1 Cap is right: This can be achieved by swapping 2 caps and the ways you can do that are $\frac{n!}{(n-k)!k!}$ which are the combinations (it doesn't matter if you swap A with B or B with A) of 2 objects from a set of 3 $\frac{3!}{(3-2)!2!}$ = 3

Thus, from the number of ways to order the caps, we have to subtract the ways that make some caps remain in the right place:

6 - 1 - 3 = 4

Total 3! = 6 ways to put on caps(permutations).

3 ways to get at least 1 cap on the right bottle.

3!/3 = 2

There are only two ways to select the different color on all the three bottles.

Total permutation to put cap = 3!=6

When all cap put on correct position =1

Number of permutation if only one cap at correct position =3

Hence no cap at correct position = 6-(3+1)=2

Using the derangement formula = $3!(\frac{1}{1} - \frac{1}{1} +\frac{1}{2} - \frac{1}{6} = 6(\frac{1}{3}) = 2$

Using permutation and computation concept..... * If there are n objects which fits in n different places such that only one arrangement is correct. then no of possible ways in which above objects can be placed in wrong order = n! *(1-1/1!+1/2!-1/3!...............1/n!)).

here n=3.

so no of possible ways=2

Since the numbers are small we could use our intuition, but here is a different way.

Imagine each bottles are set on the vertices of a triangle.

For the caps not match, either, the caps have changed their position to the next bottles clockwise once (or have moved anticlockwise twice) or the caps have changed positions anticlockwise once . ( or have moved clokwise twice)

Note: repeating a pattern 3 times would result in no movement. More than 3 would fall into one of the first 2 categories.

Therefore 2 ways.

If we want to use factorial:

The total possible no. of ways for the caps to be arranged are 3! = 6

The no. of ways for the bottle to match: 1. All of them match 2. One of them match.

Note: matching 2 of them at once will result in the third one matching eventually.

For (2) there are three caps, so three ways, so a total of 4 ways ( including (1)).

So (total number of ways - no. of ways for matching = no. of ways for the caps not to match their bottles=2)

Here is a solution that involves only basic math:

The total number of combination for the three caps is 3!=1 2 3=6. Let the caps be a,b,c and the bottles A,B,C (so that cap a is bottle A's cap etc)

So we have these combinations:

A-a, B-b, C-c
Obviously this is the original correct cap placement so the combinations we accept are now 5

A-b, B-a,C-c The cap on C is matching so the combinations are now 4

A-c, B-b, C-a The cap on B is matching. Combinations 3

A-a, B-c, C-b The cap on A is matching. Combinations 2

A-c, B-a, C-b Acceptable

A-b, B-c, C-a Acceptable

Finally only 2 of the 6 feasible combinations of the caps are acceptable.

Let the Bottle Caps be A ,B ,C probability

2 on the first bottle

1 on the second

1 on the third

2 x 1 x 1 = 2

Cap from bottle A can be on bottle B or C. Positions of the other caps will be impose by the problem criteria.

Statistically, the answer would be:

3! in order of shooting the bottles (in any order)

2 for each bottle taking only 2 other colors, which is this given question

9 for 3 different bottles times 3 different cap colors