A basketball team scored 31 points. The points were made up of three pointers (3 points), field goals (2 points), and foul shots (1 point). How many different ways could the team have scored 31 points?How do you solve for the number of different ways to score 31 points in a basketball game?
Let's call the various shots A (3 points), B (2 points), C (1 point)
And we can use the notation (a, b, c)
to mean a A's (3 pts each), b B's (2 pts each), c C's (1 pt each).
You can have from 0-10 A's.
With 10 A's
(10, 0, 1) is the only possibility
If there are 9 A's,
(9, 2, 0) (9, 1, 2) (9, 0, 4) ... 3 ways
For 8 A's
(8, 3, 1) (8, 2, 3) (8, 1, 5) (8, 0, 7) ... 4 ways
For 7 A's
(7, 5, 0) so 6 ways (b = 5, 4, 3, 2, 1, 0)
For 6 A's
(6, 6, 1) so 7 ways (b = 6, 5, 4, 3, 2, 1, 0)
and the pattern will continue
Because 31 is odd, it jumps whenever a is odd.
A's: . 10 . . 9 . 8 .. .7 . 6 . .5 .. 4 .. . 3 . . 2 . . 1 .. 0 . . A's
ways: . 1 + (3+4) + (6+7) + (9+10) + (12+13) + (15+16) ways
96 ways.
.How do you solve for the number of different ways to score 31 points in a basketball game?
If the order of the scoring matters, (so 2+1+1 is different from 1+2+1 is different from 1+1+2 in counting the scoring 4 points) then the formula for how many ways to score n points is the nearest integer to:
a^n * (4a^2+3a+2)/(7a^2+4a+3)
where a is the real root of x^3-x^2-x-1=0 (so a ~ 1.84)
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How do you solve for the number of different ways to score 31 points in a basketball game?The advanced way to do this is to use something called generating functions.
First, note that (1+x+x^2+...)(1+x^2+x^4+x^6...)(1+x^3+x^鈥?br>
is:
sum(i=0,oo) a_i x^i
where a_i is the number of ways to score i points.
But the terms in the above multiplication are geometric expressions, so:
sum a_i x^i = 1/[(1-x)(1-x^2)(1-x^3)]
Mulitply both sides by (1-x)(1-x^2)(1-x^3) = 1-x-x^2+x^+x^5-x^6, and recombine terms, and you get:
1 = sum(i=0,oo) x^i * (a_i - a_(i-1) - a_(i-2) + a_(i-4) + a_(i-5) - a_(i-6))
where we set a_i=0 for i%26lt;0.
Then a_0 = 1, and for i%26gt;0;
and a_i = a_(i-1) + a_(i-2) - a_(i-4) - a_(i-5) + a_(i-6)
This gives you a recursive definition of a_i, and you get the following values:
i a_i
1 1
2 2
3 3
4 4
5 5
6 7
7 8
8 10
9 12
10 14
11 16
12 19
13 21
14 24
15 27
16 30
17 33
18 37
19 40
20 44
21 48
22 52
23 56
24 61
25 65
26 70
27 75
28 80
29 85
30 91
31 96
32 102
33 108
34 114
35 120
36 127
37 133
38 140
39 147
40 154
41 161
42 169
43 176
44 184
45 192
46 200
47 208
48 217
49 225
50 234
There's a closed form solution that you can write as:
a_N = 25/72 + N/2 + N^2/12 + r_N,
where r_N cycles through the values:
r_0 = 47/72
r_1 = 5/72
r_2 = 23/72
r_3 = 29/72
r_4 = 23/72
r_5 = 5/72
r_(n+6) = r_n
In particular, since all of these are values between 0 and 1, exclusive, you can just compute 25/72 + N/2 + N^2/12 and round up to the next integer.
First, don't expect a very fancy method. This is sort of like ways of making $1 in change with 16 coins. If you call the scoring numbers, T,. G and F for three points, goals and fouls respectively then.....
3T + 2G + F = 31.
If F=1, T=10 max.
If F=1, G=15 max, and
if T and G=0, F=31.
So you have a lot of "wiggle room" to come up with combos to score 31 points.
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