YAMP (Yet Another Math Puzzle)

[cyclops]

I checked the tip in the book and it said to prove the formula for G as we found it. So this brings us not any further.

Next I tabled F(15,k,r):


C(n,k)↓ k↓ r→ 1 2 3 4 5 6 7 8
105 2 14 13 12 11 10 9 8 7
455 3 169 121 81 49 25 9 1 0
1365 4 870 369 111 15 0 0 0 0
3003 5 2541 441 21 0 0 0 0 0
5005 6 4795 210 0 0 0 0 0 0
6435 7 6399 36 0 0 0 0 0 0
6435 8 6434 1 0 0 0 0 0 0
5005 9 5005 0 0 0 0 0 0 0
3003 10 3003 0 0 0 0 0 0 0
1365 11 1365 0 0 0 0 0 0 0
455 12 455 0 0 0 0 0 0 0
105 13 105 0 0 0 0 0 0 0
15 14 15 0 0 0 0 0 0 0
1 15 1 0 0 0 0 0 0 0

Also of little help. So I am about to give up. I will publish the answer from the book shortly unless someone attacks this Bastille.

[perceval]

I checked the tip in the book and it said to prove the formula for G as we found it. So this brings us not any further.

Next I tabled F(15,k,r):


C(n,k)↓ k↓ r→ 1 2 3 4 5 6 7 8
105 2 14 13 12 11 10 9 8 7
455 3 169 121 81 49 25 9 1 0
1365 4 870 369 111 15 0 0 0 0
3003 5 2541 441 21 0 0 0 0 0
5005 6 4795 210 0 0 0 0 0 0
6435 7 6399 36 0 0 0 0 0 0
6435 8 6434 1 0 0 0 0 0 0
5005 9 5005 0 0 0 0 0 0 0
3003 10 3003 0 0 0 0 0 0 0
1365 11 1365 0 0 0 0 0 0 0
455 12 455 0 0 0 0 0 0 0
105 13 105 0 0 0 0 0 0 0
15 14 15 0 0 0 0 0 0 0
1 15 1 0 0 0 0 0 0 0

Also of little help. So I am about to give up. I will publish the answer from the book shortly unless someone attacks this Bastille.

mm to me it is solved:

f(n,k,r)=[C(n-(k-1)(r-1),k)-C(n-r(k-1),k)]/C(n,k)

The fact that it does not simplify to something nice is irrelevant imho.
computation wise, the k! does simplify so your are left with 3 mult of k terms, 1 substraction and one div which is simple (for a computer)

[cyclops]

Here is the solution in the book. As you see the result is the same as ours. Their derivation is a bit more elegant because they don't use multicombinations.
Where they write f_r(n,k) we wrote G(n,k,r).

Math30a.jpg (35.3 KiB) Viewed 5394 times

Math30b.jpg (42.4 KiB) Viewed 5394 times

[perceval]

Here is the solution in the book. As you see the result is the same as ours. Their derivation is a bit more elegant because they don't use multicombinations.
Where they write f_r(n,k) we wrote G(n,k,r).

book solution is elegant indeed