Iceberg
A task is the tip of a learning iceberg. There is always more to a task than is recorded on the card.


Questions 1 and 2 relate to four available flowers and can be tackled by young children. Four different flowers in the set  A, B, C, D  and the students are asked to find ways to make up bunches of 1, 2, 3 and 4.
 Bunches of 1: A ... B ... C ... D ... 4 ways
 Bunches of 2: AB ... AC ... AD ... BC ... BD ... CD ... 6 ways
 Bunches of 3: ABC ... ABD ... ACD ... BCD ... 4 ways
 Bunches of 4: ABCD ... 1 way
There is something of a rhythm to these solutions and if we suggest that there is also the choice of not picking a bunch at all  choosing a bunch of zero  then there is one way to do that  and a pattern.
 Bunches of 0: 1
 Bunches of 1: 4
 Bunches of 2: 6
 Bunches of 3: 4
 Bunches of 4: 1
So the total number of ways to make bunches with just 4 flowers = 16.
Question 3
The double line across a task card is traditionally used to indicate an increase in the level of difficulty. So although the first part of this card is a challenge for young children, it is likely that question 3 would be too much of a challenge. However these students  perhaps years 2, 3, 4  can still be extended with challenges such as:
 Explore the number of ways to make bunches if there was only ...1 ... or 2 ... or 3 flowers to choose from.
 Suppose there were more than one of each flower in the field. Explore the number of ways to make bunches if the children can put more than one of each flower in a bunch.
Exploring the first of these options (the simpler cases) gives:
1 Flower
Bunches of 0: 1
Bunches of 1: 1
Total = 2 
2 Flowers
Bunches of 0: 1
Bunches of 1: 2
Bunches of 2: 1
Total = 4 
3 Flowers
Bunches of 0: 1
Bunches of 1: 3
Bunches of 2: 3
Bunches of 3: 1
Total = 8

Not only are these simpler cases a worthwhile investigation for younger students, they are also essential to helping older students get the most value from question 3. Trying to record all the ways of making bunches of 4 from 6 flowers requires care and perseverance. Many students will rise to the challenge. These photos show possible ways to organise the count.
The photo above is applying the strategy of making a list or table. The one to the right is making model (or perhaps a physical list) and both are breaking the problem into smaller parts. Both also use a reference set of flowers. 

The photo to the right shows all the ways to make bunches of four from a set of just six possible. It also raises the question of creating a rule / recipe / algorithm  the sort of instructions a computer would need  to faultlessly produce the full list (without the sledge hammer method of trying every possible combination and ignoring the ones that don't fit the problem).
 Can the students create a set of instructions for writing out the full list of combinations resulting from starting with a set of 6 objects and choosing of 4 of them?
Note: If the students would like to know the mathematical shorthand for this mouthful it might be time to introduce one of the three common methods: _{6}C_{4} or ^{6}C_{4} or (^{6}_{4}).
Again breaking the problem into smaller parts might help, this time focussing on the system used to write out the combinations. For example, starting with 6 flowers (A, B, C, D, E, F) and choosing...
Bunches of 0:
 Choose one to start ... which can only be done by not doing it.
 1 way
Bunches of 1:
 Choose one to start and join with no other because you have a bunch of 1  A
 Choose the next one not yet chosen to start a bunch and join with no other  B
 ...continue until there are none left unused as the starting one
 6 ways
Bunches of 2:
 Choose one to start and join with one other  A x 5
 Choose the next one not yet chosen to start a bunch and join with one other to the right in the reference list  B x 4
 Choose the next one not yet chosen to start a bunch and join with one other to the right in the reference list  C x 3
 Choose the next one not yet chosen to start a bunch and join with one other to the right in the reference list  D x 2
 Choose the next one not yet chosen to start a bunch and join with one other to the right in the reference list  E x 1
 Choose the next one not yet chosen to start a bunch and join with one other to the right in the reference list  F has none left to join with
 15 ways
Bunches of 3:
 Choose one to start, join with the next to the right to make a pair, then join with one other to make 3  AB x 4
 Using the same start, join with the next to the right to make a pair, then join with one other to make 3  AC x 3
 Using the same start, join with the next to the right to make a pair, then join with one other to make 3  AD x 2
 Using the same start, join with the next to the right to make a pair, then join with one other to make 3  AE x 1
 Choose the next one not yet chosen to start a bunch, join with the next to the right to make a pair, then join with one other to make 3  BC x 3
 Choose the next one not yet chosen to start a bunch, join with the next to the right to make a pair, then join with one other to make 3  BD x 2
 Choose the next one not yet chosen to start a bunch, join with the next to the right to make a pair, then join with one other to make 3  BE x 1
 Choose the next one not yet chosen to start a bunch, join with the next to the right to make a pair, then join with one other to make 3  CD x 2
 Choose the next one not yet chosen to start a bunch, join with the next to the right to make a pair, then join with one other to make 3  CE x 1
 Choose the next one not yet chosen to start a bunch, join with the next to the right to make a pair, then join with one other to make 3  DE x 1
 20 ways
Using this algorithm again to make triples first and then joining them with one other gives
Bunches of 4:
 ABC x 3
 ABD x 2
 ABE x 1
 ACD x 2
 ACE x 1
 ADE x 1
 BCD x 2
 BCE x 1
 BDE x 1
 CDE X 1
 15 ways
Bunches of 5:
 ABCD x 2
 ABCE x 1
 ABDE x 1
 ACDE x 1
 BCDE x 1
 6 ways

Bunches of 6:
A grand total of 64 ways, which confirms the earlier pattern for the total of ways for any starting number of flowers.
Writing out the data like this highlights the rhythm of each set of solutions and makes it easier to predict missing data.
In this context, the 1 at the apex is essentially for completion of the pattern although perhaps we could think of it as not going to the field to pick flowers at all. There is only one way to do that because we either go or we don't.

However we can also add an extra column to each row to record the total of each as a power of two. If we did then working backwards through that column would lead to the conclusion that 2^{0} = 1. In fact, mathematicians agree that the only definition of 2^{0} which makes sense is that it equals 1. This is also true for any other number raised to the power of zero.
This triangle arrangement is called Pascal's Triangle after the mathematician Blaise Pascal. It has several uses in mathematics.
Challenge
This last section of the card raises the problem to yet another level. As we have seen, and hopefully as you have discovered yourself before reading these notes, counting combinations is tricky. The 'computer' instructions above go some way towards a generalisation for recording the possible ways of making bunches, but they don't tell us in advance the number of ways that there will be. The challenge is asking for a way to predict that answer for any number of flowers (n) and any group size (r).
One was to do this is to start with the simpler problem of counting ordered selections  as if the flowers were going to be placed one at a time in numbered vases along the mantelpiece. We still assume only one of each flower, so no repeats. This much easier to count:
 n choices for the flower in the 1st vase, then, for each of these choices...
 (n  1) choices for the flower in the 2nd vase then, for each of these pairs of choices so far...
 (n  2) choices for the flower in the 3rd vase then, for each of these triples of choices so far...
 ... this continues and the choices dwindle until in the third last vase there are only...
 3 flowers to choose from, then in the second last vase only...
 2 flowers to choose from, then in the last vase only...
 1 flower to choose from.
The total of these ordered arrangements = n x (n  1) x (n  2) x ... x 3 x 2 x 1. The students might like to know the mathematicians symbol for this, which is n! (n factorial) because it surely saves a lot of writing.
So, counting ordered arrangements is easy, but it's not the problem we are trying to solve. Our bunches aren't ordered and even if they were we only want to use r vases, not n.
The restriction of counting ordered arrangements into r vases is also easy to count. We simply stop using flowers when the rth vase has a flower in it. That doesn't change the fact that the count until then begins the same, that is, n x (n  1) x (n  2) x ... The question is only how much of the end of the multiplication isn't used. Considering a few cases from above soon makes that clear. If we were making ordered selections:
 starting with 6 and using 4 vases, the multiplication would be 6 x 5 x 4 x 3 and the unused part would be 2 x 1.
 starting with 6 and using 3 vases, the multiplication would be 6 x 5 x 4 and the unused part would be 3 x 2 x 1.
 starting with 6 and using 2 vases, the multiplication would be 6 x 5 and the unused part would be 4 x 3 x 2 x 1.
The examples suggest that the unused part is (n  r)! and that if divide the full multiplication by that, we will be left with the count for the number of ordered arrangements in r vases. That is:
^{n!} / _{(n  r)!}
So we are closer to the generalisation asked for, but our flowers aren't ordered. Mother will put her bunch into one vase. We have several times too many in the count at the moment and we can work out the number of times too many. Look at the case above again.
 We have arranged 4 flowers in 4 vases which can be done in 4 x 3 x 2 x 1 ways, so the count is that many times too big.
 We have arranged 3 flowers in 3 vases which can be done in 3 x 2 x 1 ways, so the count is that many times too big.
 We have arranged 2 flowers in 2 vases which can be done in 2 x 1 ways, so the count is that many times too big.
In general then the count in the ordered selection above is r x (r  1) x (r  2) x ... 3 x 2 x 1 = r! times too big if we want a bunch, or unordered selection (also called a combination). So, if we divide the ordered selection count by r! that will be the count for the unordered selection.
If we start with n flowers and make an unordered selection of r flowers, the number of ways to do this is:
[^{n!} / _{(n  r)!}] divided by r!
which simplifies to:
^{n!} / _{(n  r)! r!}
Extensions
 For any ^{n}C_{r} the total number of combinations is a power of 2. Why should that be?
 For any ^{n}C_{r} the total of the number of flowers used in each colour is the same across the set of possibilities. Why should that be?
To clarify 2, if necessary, count the number or each colour in the photo above. Ten of each colour have been used to make the complete set of possibilities.
