Maths homework - combinations

[swimchick1980]

DD (10) has homework. There are 4 black beads and 4 white beads. How many different combinations are there for making a bracelet. I don't know where to start other than by laying them all out. The rest of her homework was squaring things and square root. Can anyone help? 🙏

Do you have to use all eight beads?

If so, you could start by assuming you start with four black beads. Then the next four are white, so that's one combination.

Then start with 3 black beads. You could then have a black one at position 6, 7, or 8.

If you started with 2 black beads (and a white in position 3), then work out where the other two black beads could go (e.g., 4/5, 4/6, 4/7, 4/8, 5/6, 5/7, 5/8, 6/8)

Then start with 1 black bead (white in position 2) and imagine first that there's a black in position 3, and then work out all the other possibilities like the previous step. then repeat it with the second black in position 4, and so on.

If you can have fewer than 8 beads, then you'll have to repeat the whole process with different numbers, but that will incresae the number of combinations hugely, so I would assume they mean using all the beads if it doesn't say. Also I am assuming that there is a start and an end to the bracelet! Otherwise there would be fewer combinations as if it's circular, there will be lots of repeats

It's just about coming up with a systematic way of listing the possibilities for this age; they don't need to find a rule or a formula to do it.

I find that it's useful to keep things the same as much as possible and just change the last element, until you've done all those combinations. Then just change the second last element, and work through those, and so on.

Don't just try to come up with random combinations until you have them all.

There are of course rules and formulas that you can look up, but not helpful for this age.

https://brainly.com/question/32787089 Check here , exactly your question

If you can't open , here is the answer " To determine the number of different bracelets that can be made with 4 white beads and 4 black beads, we can use the concept of combinations.

First, let's consider the number of ways to arrange the 8 beads in a straight line without any restrictions. This can be calculated using the formula for permutations, which is 8! (8 factorial).

However, since we are making bracelets, the order of the beads in a circular arrangement doesn't matter. We need to account for the circular symmetry by dividing the total number of arrangements by the number of rotations, which is 8 (since a bracelet can be rotated 8 times to yield the same arrangement).

Therefore, the total number of distinct bracelets can be calculated as:
Number of bracelets = (Number of arrangements) / (Number of rotations)
= 8! / 8

Simplifying this expression, we get:
Number of bracelets = (8 7 6 5 4 3 2 * 1) / 8
= 7 6 5 4 3 2 1
= 7!

Using the formula for factorials, we can calculate:
Number of bracelets = 7! = 7 6 5 4 3 2 1 = 5040

Therefore, there are 5040 different bracelets that can be made with 4 white beads and 4 black beads."

Here's an example of how you could write them out, showing how certain things repeat (obviously you could do the ordering in lots of different but logical ways). I've here ordered them by how many blacks in the first half, and then with each of those, you have several ways to arrange the number of blacks in the second half).

bbbb wwww (four blacks in first group, and only one way to do this)

bbbw bwww (three blacks in first group, and there are 4 ways to do this)
bbbw wbww
bbbw wwbw
bbbw wwwb

bbwb bwww
bbwb wbww
bbwb wwbw
bbwb wwwb

bwbb bwww
bwbb wbww
bwbb wwbw
bwbb wwwb

wbbb bwww
wbbb wbww
wbbb wwbw
wbbb wwwb

bbww bbww (two blacks in first group, and there are six ways to do this)
bbww bwbw
bbww bwwb
bbww wbbw
bbww wbwb
bbww wwbb

bwbw bbww
bwbw bwbw
bwbw bwwb
bwbw wbbw
bwbw wbwb
bwbw wwbb

bwwb bbww
bwwb bwbw
bwwb bwwb
bwwb wbbw
bwwb wbwb
bwwb wwbb

wbbw bbww
wbbw bwbw
wbbw bwwb
wbbw wbbw
wbbw wbwb
wbbw wwbb

wbwb bbww
wbwb bwbw
wbwb bwwb
wbwb wbbw
wbwb wbwb
wbwb wwbb

wwbb bbww
wwbb bwbw
wwbb bwwb
wwbb wbbw
wwbb wbwb
wwbb wwbb

bwww bbbw (one black in first group, and there are 4 ways to do this)
bwww bbwb
bwww bwbb
bwww wbbb

wbww bbbw
wbww bbwb
wbww bwbb
wbww wbbb

wwbw bbbw
wwbw bbwb
wwbw bwbb
wwbw wbbb

wwwb bbbw
wwwb bbwb
wwwb bwbb
wwwb wbbb

wwww bbbb (no blacks in first group, and there is 1 way to do this)

You might recognise the numbers in each 'group' of combinations:
1, 4, 6, 4, 1 (pascal's triangle)

This answer is wrong though. It would be right if there were 8 different beads.

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@juliaxxl80 Incorrect. This assumes that black bead 1 is different from black bead 2 etc.

Unfortunately, I suspect that this is one where the teachers expect you to write out all the combinations. Which isn't proper maths to me. But there you go.

Yes I was also assuming that the black beads were indistinguishable, and so were the white beads.

And I was assuming that the bracelet had a start and end to it where a clasp would be, rather than, say, on an elastic with no start or end.

And I assumed you needed to use all eight beads, and nothing else.

Any of those assumptions could change the answers, so you probably want to note that in whichever solution you come up with.

But I would not go into formulas really with a 10-year old, but instead looking at things like the fact that the patterns repeat in predictable ways; you don't have to write them all out each time (e.g., I wasn't going to write out the second half of each of my sets each time, but was just going to say 'like the previous ones', but I could easily copy them so I did in the end). It's quite helpful when a child is writing them out and realises the repeats, and sees that they could just copy or ditto something previous, as it helps them work out what the pattern is.

1 16 36 16 1

My assumptions were the same, except without the clasp, so there's an extra task of working out what that means for the result.

I suspect I may have missed a couple, but going from the long list of 70, and trying to eliminate rotations, I have cut it down to 12. It becomes easier to think of various groupings of 1, 2, 3, and 4 of each colour and what comes in between them etc.

Of course, if they are actually idential beads, and totally round or something, there's always the possibility that you might be able to put it on upside-down, and thus occasionally get the same pattern again as one of the others, so that would reduce it further. But I'm getting a bit lost working it out!

This is hw for a 10 year old.
Don't overthink it.

Start with bbbbwwww
Then bbbwbwww, bbbwwbww, bbbwwwbw, bbbwwwwb
and go on from there.

If the child is keen, explore which combinations could be considered 'the same' without a clasp etc, but not otherwise.

Just musing now:
How many 8 bit binary numbers are there using exactly 4 zeros and 4 ones? Because that would be the same number.

I cannot believe this is for Year 4!

Absolutely! I thought, it was for Year 10 .

Has she been doing binary at school?

When accounting for both rotational and reflectional symmetry, there are 10 unique ways to arrange a bracelet with 4 black beads and 4 white beads.

This result aligns with the combinatorial problem where symmetry plays a significant role, and it often helps to use combinatorial tools or group theory for such problems.

Beautifully over and above @BlossomToLeaves , that Brainly answer is outrageously wrong, can see why you were prompted to create the full list.

Half the number, to remove the mirror images, and you get 35 possible bracelets. (if the bracelet has a clasp)

Can't believe I woke up at 5am and thought 'but what if it was on elastic?' 😩

Anyone who has developed a taste for maths - check out The Number Devil by Hans Magnus Enzensberger, he is slightly obsessed with Pascal's triangle.

I can only find 8 😩

I think the difference might be in deciding which types of symmetry to accept. For example, when I first went through to eliminate different rotations of the pattern (as on an elastic), I was still assuming that there was a right way up, and a right way out, which means that mirror reflections would be different bracelets.

So in the diagram above, the second pattern in the first row, for example, which has three blacks followed by three whites, would be a different pattern to one that had three whites followed by three blacks, because unless you turned the bracelet upside down to put it on, those would count as different to me. You could rotate it around your wrist, but you still wouldn't change 3 blacks then 3 whites to the other way around, without flipping it over. Similarly for the first pattern in the third row, with two whites followed by two blacks. I think you would have a different pattern with two blacks followed by two whites.

When I did that on my original 70 patterns, I cut it down loads, to I think 12, but probably should have been only 10, as I think I probably still had some doubles in there. I think it's probably the eight shown above, plus the two I've just described in their alternate forms. I can only see those two at the moment that could have the reverse forms, but it's possible I've missed one again.

And as always, it depends a lot on what is considered a 'unique' arrangement!

I agree.
All the other are symmetrical.

This sounds like one of those questions that is a lot more complicated than the teacher realises.

That's really hard to do. I think a word with the teacher might be a good idea.

I think for the minute I'd tell DD to do it as a circle with no defined positions, because you'll get into horrendous numbers very quickly otherwise.

The ten above looks right to me with those assumptions I think, but I'm not completely confident (and I'm usually quite good at logic and maths!).

If you wanted to do it with the combination formula I think it would be

C(8, 4) = 8! / (4! * 4!) = 70

But that doesn't account for the fact that a lot of those would look exactly same on your wrist!

I think they are 10.
The teacher just wants them to think methodically.
Differences in answers can be discussed in class.
This isn't A level.

This is such a great example of a homework problem because it can be tackled at loads of different levels. I'm glad I'm not the only person still thinking about it.

Having to account for both rotational symmetry and mirror symmetry is breaking my brain!

As the question says 'different bracelets' I think you have to take all forms of symmetry into account. If you had a jumble of bracelets in a drawer and you picked two out, you'd count them as the same if you could make them look the same by turning them round or turning them over.

With that in mind I came at this by asking 'what's allowed to touch'. If you can only have different coulors touching, there's only 1 way. Equally if you can only have 4 colours touching there's only 1 way.

Pairs are more difficult. If it's all pairs, what possible combinations are there? If there are exactly 4 pairs - 1 way. 3 pairs and 2 singles is not possible because the leftover singles would make another pair. 2 pairs and 4 singles - 1 way because the singles must be separated, otherwise they make more pairs. 1 pair and 6 singles - not possible because if you start with BB and then go W B W B... the final one will create a triple. So only 2 ways to have pairs, either pair/pair/pair/pair or pair/single/single/pair/single/single.

If 3 are allowed to touch then they can be separated from their remaining single by either two pairs or a 1 and a triple. If there are 2 pairs, there are 2 variations depending on whether the triple is black or white. If there is a 1 and a triple there will be two triples and there's only 1 way to do this.

So altogether that comes to 8. This aligns with @NerdWhoEatsMedlar 's diagrams with the amendments made by @BlossomToLeaves but having thought it out like this, I'm more satisfied that there isn't any double counting or omission. I'd be very interested to see how @SugarHorseSpooks's solution compares.

Doggo will now be having a shorter walk than expected because I've spent far too long on this!