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# Partitions of 20p watch

1. Hey I have started to work through Stephen Siklos' book 'Advanced Problems in Mathematics'.

For Problem 2: Partitions of 10 and 20, the author mentions an incorrect way of approaching the problem:

"Since different parts of STEP questions are nearly always related, you might be led to believe that the result of the second part follows from the first: you divide the required twenty pence into two tens and then use the result of the first part (i.e. 11 ways to make 10p) to give the number of ways of making up each 10. This would give an answer of 66 (why?) plus one for a single 20p piece."
Well I have highlighted my question. Why would this line reasoning give us 66 (+ the single 20p piece)I would've thought that the answer for this line of reasoning is: 11*11 = 121.Can someone please explain to me how 66 is obtained.
2. If I point out that 66 is the sum of he first 11 integers, does that give you any ideas?
3. this should help....

https://www.openbookpublishers.com/h...apters/P2.html
4. @Pangol: No it doesn't. How am I using the previous part?
5. (Original post by FXLander)
@Pangol: No it doesn't. How am I using the previous part?
We're looking at partitioning 20p by treating it as two lots of 10p. Let's call these two halves 10A and 10B.

Suppose we partition 10A using one of the 11 partitions we already know about. Then 10B can be partitioned in 11 ways - we have 11 ways so far.

We then partition 10A using another one of the 11 partitions. We could now partition 10B using the 11 ways, but one of these ways will be the same as what we have just done, so we really only have 10 ways of partitioning 10B. We therefore have 11 + 10 ways so far.

Carry on like this and we will have 11 + 10 + 9 + ... + 1 = 66 ways altogether.
6. (Original post by Pangol)
We're looking at partitioning 20p by treating it as two lots of 10p. Let's call these two halves 10A and 10B.

Suppose we partition 10A using one of the 11 partitions we already know about. Then 10B can be partitioned in 11 ways - we have 11 ways so far.

We then partition 10A using another one of the 11 partitions. We could now partition 10B using the 11 ways, but one of these ways will be the same as what we have just done, so we really only have 10 ways of partitioning 10B. We therefore have 11 + 10 ways so far.

Carry on like this and we will have 11 + 10 + 9 + ... + 1 = 66 ways altogether.
When you say "Suppose we partition 10A using one of the 11 partitions" that I don't understand. Why is it that we have 11 ways for 10B, but randomly make a choice of 'one of the partitions we know about' for 10A, I find it difficult to visualise what is going on...
7. (Original post by FXLander)
When you say "Suppose we partition 10A using one of the 11 partitions" that I don't understand. Why is it that we have 11 ways for 10B, but randomly make a choice of 'one of the partitions we know about' for 10A, I find it difficult to visualise what is going on...
Sorry if I wasn't clear - I was suggesting that we systematically partition 10A in each of the 11 ways that we know about. Let's call these 11 partitions P1, P2, ..., P11.

When we partition 10A using P1, we can partition 10B using P1, P2, ..., P11. So, 11 ways.

When we partition 10A using P2, we can partition 10B using P1, P2, ..., P11. But partitioning 10A with P2 and 10B with P1 is the same partition of the entire 20p as partitioning 10A with P1 and 10B with P2, which we have already done. So, we only have an additional 10 ways this time, which together with what we have already done, gives 11 + 10 partitions in total.

When we partition 10A using P3, we can partition 10B using P1, P2, ..., P11. But we have effectively already done a combination of P3 with P1 and P3 with P2, so we only have an additional 9 ways this time, which together with what we have already done gives 11 + 10 + 9 partitions in total.

And so on, until we have 11 + 10 + 9 + ... + 2 + 1 = 66 partitions in total.

As is noted in the book, there are other duplications that have to be taken into account as well so that the real answer is less than 66, but this is the reason why we might expect there to be 66 ways (plus the one way of just using one 20p piece).
8. (Original post by Pangol)
Sorry if I wasn't clear - I was suggesting that we systematically partition 10A in each of the 11 ways that we know about. Let's call these 11 partitions P1, P2, ..., P11.

When we partition 10A using P1, we can partition 10B using P1, P2, ..., P11. So, 11 ways.

When we partition 10A using P2, we can partition 10B using P1, P2, ..., P11. But partitioning 10A with P2 and 10B with P1 is the same partition of the entire 20p as partitioning 10A with P1 and 10B with P2, which we have already done. So, we only have an additional 10 ways this time, which together with what we have already done, gives 11 + 10 partitions in total.

When we partition 10A using P3, we can partition 10B using P1, P2, ..., P11. But we have effectively already done a combination of P3 with P1 and P3 with P2, so we only have an additional 9 ways this time, which together with what we have already done gives 11 + 10 + 9 partitions in total.

And so on, until we have 11 + 10 + 9 + ... + 2 + 1 = 66 partitions in total.

As is noted in the book, there are other duplications that have to be taken into account as well so that the real answer is less than 66, but this is the reason why we might expect there to be 66 ways (plus the one way of just using one 20p piece).
I understand it now. Thanks for the clear explanation!

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