Quandaries
and Queries 

Name: Claire Student Level: Secondary Topic: Algebra and Number patterns We need to be able to discover the nth term in a pattern of numbers and explain how we did it. We have arrived at the formula 1+2+3+...n=n(n+1)/2 for trios where a trio series is; 3 has the following trio (1,1,1) 4 has the following trios (2,1,1), (1,2,1), (1,1,2) 5 has the following trios (3,1,1), (1,3,1) (1,1,3) (1,2,2) (2,2,1) (2,1,2) So the series is as follows Term Value Trios 1 3 1 2 4 3 3 5 6 And so on. Our problem is working out the formula for quads where a quad series is as follows 4 has the following quad (1,1,1,1) 5 has the following quads (1,1,1,2) (1,1,2,1) (1,2,1,1) (2,1,1,1) etc etc 

Hi Claire, You need to know a bit about combinations, the number of ways of choosing k of n distinct objects, nCk. The problems are equivalent to the following: Suppose that you have n identical lollipops and and m children come to your door; in how many ways can you hand out all of the lollipops such that each child gets at lest one lollipop? For example, with 5 lollipops and 3 children you can hand them out as (3,1,1), (1,3,1) (1,1,3) (1,2,2) (2,2,1) (2,1,2) where in the first case we have the first child gets 3 and the others get one each, and so on. Suppose that you had 10 lollipops and 3 children, how can you count the ways? First, why not give each child one so that we don't have to worry about someone being left out. This leaves us with 7 lollipops to distribute to three children but we no longer have to worry about the someone getting none condition. Think of 7 x's and 2 's. For each ways of lining up these symbols we have a distribution of the lollipops  for example xxxxxxx means give the first child 2, the second 4 and the third 1; xxxxxxx means give the 3, 0 and 4 respectively, and so on. Note in these two examples (because we already gave them 1 each) the children get (3,5,2) and (4,1,5). In how many ways can we line up these 9 symbols? If we choose 2 spots for the 's then the x's must go n the other spots so we can do this in "9 choose 2" = 9C2 = 9(8)/2 ways. In general for n lollipops and 3 children you will get [(n3)+2] choose 2 = (n1)(n2)/2 ways. Suppose that you had 10 lollipops and 4 children, how
can you count the ways? First, why not give each child one so that
we don't have to worry about someone being left out. This leaves us
with 6 lollipops to distribute to 4 children but we no longer have
to worry about the someone getting none condition. Think of 6 x's and
3 's. In how many ways can we line up these 9 symbols? If we choose
3 spots for the 's then the x's must go n the other spots so we can
do this in "9 choose 3" =
9C3 = 9(8)(7)/2 ways. As above, xxxxxx gives us 2, 1,1,2 and adding
in the first lollipops that we gave out we have the solution (3,1,2,3),
etc. In general for n lollipops and 4 children you will get [(n4)+3]
choose 3 = (n1)(n2)(n3)/3! = (n1)(n2)(n3)/6 ways. Penny


