On Friday I posted this puzzle….

My friend has 4 pieces of jewelry – a ring, a necklace, a bracelet and a brooch. Given that she can wear any number of the objects when she goes out, how many combinations of her jewelry does she have to choose from?

If you have not tried to solve it, have a go now. For everyone else, the answer is after the break.

She can wear 4 combinations of 1 piece, 6 of 2, 4 of 3, and 1 of 4 – making 15 different combinations in all. Did you solve it?

I have produced an ebook containing 101 of the previous Friday Puzzles! It is called **PUZZLED** and is available for the **Kindle **(UK here and USA here) and on the **iBookstore** (UK here in the USA here). You can try 101 of the puzzles for free here.

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Surely there are 16 combinations – the last being wearing no jewellery at all?

O give it a rest.

Yup, without question the answer is 16.

I say without question, what I mean is ‘without intelligent question’. The wording of the original question left no doubt that no jewellery was a valid option.

Thing is, there are clearly 12 ways of wearing two pieces, not the six claimed.

See for yourself: pick any one of the four. That’s four options. Then pick any one of the the three remaining.

4 * 3 = 12.

I agree with 16.

There are six ways for 2 pieces. You’ve double counted the options to get to 12 since order doesn’t matter.

Garry Boddard: Your method counts every option twice. For example if you list out all 12 ways you describe, you would see both “ring, necklace” and “necklace, ring” in your list. So, there’s only 6 ways to wear two pieces.

Oops. Mike already pointed out what I said.

The ‘Wiseman’ clearly states in the set up of the puzzle: “Given that she can wear any combination OF THE OBJECTS when she goes out, how many combinations OF HER JEWELRY does she have to choose from?” Come on folks, there is absolutely no suggestion that no jewelry, none, zero, nada is an option! We look for trickery when we try to solve these types of puzzles because trickery is often part of a puzzle’s premise. But so very often the design of a puzzle is in it’s simplicity. Thanks, lars the lars.

Please read the question. Absence of jewelry is not an option.

Surely – and 1 combination of none of the pieces? After all it does say “she can wear any number of the objects when she goes out” so that must include zero of them? So 16 combinations.

Yay! I got a Maths puzzle right.

I also took naught as an option and got 16 as a total.

2^4-1

What is the proof?

There are four jewelry slots. Every slot has two values, occupied or empty.

So that’s 2*2*2*2 combinations. Minus one because we don’t count all empty (not wearing any jewelry).

One necessary assumption is that order is irrelevant (ie. AB is equivalent to BA). Wearing a necklace and a ring is equivalent to wearing a ring and a necklace.

best answer!!!!

I solved it binary:

ring: yes/no (1/0)

necklace: yes/no (1/0)

bracelet: yes/no (1/0)

brooch: yes/no (1/0)

Giving options 0000 to 1111, which means there are 16 combinations… And I included the “no jewelery” as valid option.

And then I figured that this was overcomplicated as 2^4 would suffice…

Glad to see someone else solved this quickly with Binary 🙂

Yeah, That’s how I solved it. Came up with 16, foolishly because I counted the “no jewellery” option. But the idea of wearing no jewellery is a way of wearing jewellery is too silly to contemplate.

For those who think 16 is the correct answer. Please explain how 4+6+4+1+0 = 16.

4+6+4+1+1 = 16 (1 combination of 0 pieces)

As Walter say, one combination of 0 pieces is not +0. It seems a lot of us came up with 16.

For the benefit of Per:

4(options of one item) + 6(options of two items) + 4(options of three items) + 1(option of four items) + 1 (option of no items – this isn’t zero options) = 16

how do you combine nothing?

So very tired of middle school math problems being passed off as puzzles. They are neither challenging nor interesting. You can do better, Richard.

You got it wrong then, I take it.

As a middle aged + man who took calculus and statistics about 40 years ago but have forgotten most of it, I greatly enjoyed the combination problem because it made me realize I was rusty on the formula and found out what it was. Of course it’s easy to just write out all the non-repeating possibilities, but the formula is a little trickier to remember.

Here is the formula for n choose k where in this case we have 4 things (n) and choosing different combinations of k = 4, 3, 2, and 1.

n!/k! x (n-k)!

But the tricky part is you have to do it 4 times and add together.

1) n!/{k! x (n-k)!} for 4 things taken/worn 4 at a time = 4!/{4! x (4-4)!} = 1

so 1 way to wear all 4 things at same time

2) n!/{k! x (n-k)!} for 4 things taken/worn 3 at a time = 4!/{3! x (4-3)!} = 4

so 4 ways to wear 3 things at same time

3) n!/{k! x (n-k)!} for 4 things taken/worn 2 at a time = 4!/{2! x (4-2)!} = 6

so 6 ways to wear 2 things at same time

4) n!/{k! x (n-k)!} for 4 things taken/worn 1 at a time = 4!/{1! x (4-1)!} = 4

so 4 ways to wear 1 thing at same time

Adding all 4 together: 1 + 4 + 6 + 4 = 15

The problem said WEARING some combination so NOT-wearing anything at all is not included just as if there were a problem asking about flipping coins – you wouldn’t include not flipping in the results.

No Garry. I solved it.

It’s just a simple math problem. That’s the source of my disappointment. I desire a PUZZLE, not a simple math exercise.

Less than 5 seconds

I think the correct answer is 15 OR 16. Recall that the word “number” was part of the puzzle.

If you ask Google to define “number” you get definitions that include 0 and you also get:

number – definition of number by the Free Online Dictionary …

http://www.thefreedictionary.com/number

a. A member of the set of positive integers; one of a series of symbols of unique meaning in a fixed order that can be derived by counting.

You can’t have a combination of one piece! For it to be a combination there must be at least two.

In combinatorial mathematics (combinatorics) you can most certainly have a combination of one piece, whatever common sense might tell you.

Like Geodetective and others, I solved it almost immediately using the binary wear/not wear choice for each piece — much easier than Richard’s proposed solution. After that it’s really a matter of taste whether you want to call “no jewelry at all” a combination. A mathematician or logician would probably say yes; a woman who owns any jewelry would probably say no; a man who wants to enjoy going out would definitely agree with her.

Having thought about the wording a little more, it seems that you shouldn’t just rule out the “no jewellery” option, but also the four “single pieces of jewellery” option, since that can’t be thought of as a combination either. So I’m changing my answer to 11.

Those who fruitlessly argue 15 vs 16 ways to combine the jewelerry are missing the wider point as is often the case with Richard’s puzzles.

Yes there are 15 ways to wear the jewels. Yet there are an infinite number of ways not to wear them. Thus the two answers are 15 or infinity depending on your choice of picking zero as a solution.

Once again i think deeper thinking has helped clear up what has been clearly a problem for some thinkers and partial solvers.

re: one eyed Jack.

As a middle aged + man who took calculus and statistics about 40 years ago but have forgotten most of it, I greatly enjoyed the combination problem because it made me realize I was rusty on the formula and found out what it was. Of course it’s easy to just write out all the non-repeating possibilities, but the formula is a little trickier to remember.

Here is the formula for n choose k where in this case we have 4 things (n) and choosing different combinations of k = 4, 3, 2, and 1.

n!/k! x (n-k)!

But the tricky part is you have to do it 4 times and add together.

1) n!/{k! x (n-k)!} for 4 things taken/worn 4 at a time = 4!/{4! x (4-4)!} = 1

so 1 way to wear all 4 things at same time

2) n!/{k! x (n-k)!} for 4 things taken/worn 3 at a time = 4!/{3! x (4-3)!} = 4

so 4 ways to wear 3 things at same time

3) n!/{k! x (n-k)!} for 4 things taken/worn 2 at a time = 4!/{2! x (4-2)!} = 6

so 6 ways to wear 2 things at same time

4) n!/{k! x (n-k)!} for 4 things taken/worn 1 at a time = 4!/{1! x (4-1)!} = 4

so 4 ways to wear 1 thing at same time

Adding all 4 together: 1 + 4 + 6 + 4 = 15

The problem said WEARING some combination so NOT-wearing anything at all is not included just as if there were a problem asking about flipping coins – you wouldn’t include not flipping in the results.

To satisfy both sides, the answer is 15 if the lady has to wear some piece of jewelry and the answer is 16 if you the lady can choose to not wear any piece of jewelry (in that case it is easy for any idiot to add one more possibility).

Now both sides of the argument are happy.

There are 16 combinations — you omitted the “none” option. The formula for N items is simply 2^N .

The lack of sense of humor in the comment sector is flabbergasting. The grasp of irony stupendously inadequate. The linguistic proficiency at a scarily low level.

Guys, lighten up! Life is full of ambiguities, uncertainties and lines to read between.

A puzzle doesn’t need to have an elegant mathematical solution to be good. Sometimes goofy is good.

You showed one thing – it’s impossible to make a numerical joke. Sadly.

the fun of richard’s puzzles is not the solving, but of finding out why the puzzle is ambiguous or richard’s answer is wrong or incomplete.

here, there are two ambiguities: “wear any number” and “how many combinations.” the correct answer depends on whether “any number” includes 0, and whether “combinations” means that she must be wearing at least 2 pieces.

hear hear ctj,

a mundane maths problem livened up by a dodgy bit.

Why so much kerfuffle? There are 4 items, each with a yes/no option. Therefore it’s 2^4 which is 16.

I prefer nudity

Please post photo

Genial

Yes, I solved it.