Score!

Probability Level 3

In basketball, a player can score points in 3 different ways: $1$ point from the foul line, $2$ points for shots close to the basket, or $3$ points for shots far away from the basket. If a player scores $30$ points in a game, how many different ways can this be achieved?

Details and assumptions

The order of the baskets doesn't matter. For example, if the player scores 9 3-pointers and then 3 1-pointers, this is the same as scoring 3 1-pointers and then 9 3-pointers.

The answer is 91.

8 solutions

Whatever 21
Dec 2, 2013

There's only 1way of scoring 30points with only 3-point shots. 10 3-point shots.

What if there was 1 less 3-point shot? 27points will be achieved. There are 2ways of making a 3-point shot using only 1-point and 2-point shots, for making it total 30points.

. For 2 less 3-point shots (that's 6points less than 30), there are 4ways for scoring the 6points using only 1-point and 2-point shots.

For 3 less 3-point shots (that's 9points less than 30), there are 5ways for scoring the 9points using only 1-point and 2-point shots.

For 4 less 3-point shots (that's 12points less than 30), there are 7ways for scoring the 12points using only 1-point and 2-point shots.

For 5 less 3-point shots (that's 15points less than 30), there are 8ways for scoring the 15points using only 1-point and 2-point shots.

For 6 less 3-point shots (that's 18points less than 30), there are 10ways for scoring the 18points using only 1-point and 2-point shots.

For 7 less 3-point shots (that's 21points less than 30), there are 11ways for scoring the 21points using only 1-point and 2-point shots.

For 8 less 3-point shots (that's 24points less than 30), there are 13ways for scoring the 24points using only 1-point and 2-point shots.

For 9 less 3-point shots (that's 27points less than 30), there are 14ways for scoring the 27points using only 1-point and 2-point shots.

For 10 less 3-point shots (that's 30points less than 30), there are 16ways for scoring the 30points using only 1-point and 2-point shots.

total ways for scoring 30= 1+2+4+5+7+8+10+11+13+14+16=91

Does anyone have an easier way to calculate this? Is this the only way? (cause it seems quite tedious... :()

Jackal Jim - 7 years, 6 months ago

I didn't find a way to use Pick's theorem but my friend and I found another way to solve the question: (quite tedious I guess but I managed to arrive at a generalization).

Here it goes:

Mathematically, this question translates to finding the number of ordered nonnegative integers $(a,b,c)$ that satisfies the equation $3a+2b+c = 30$ . Letting $x = a+b+c, y = a+b, z = a$ , this question simplifies into finding the number of ordered nonnegative integers $(x,y,z)$ that satisfies the equations $x+y+z = 30$ and $x \geq y \geq z$ .

From here, we could use the stars and bars technique to work out the number of ordered $(x,y,z)$ that satisfies the above without condition 1 $x \geq y \geq z$ ---------- ${32} \choose {2}$ $= 496$ . BUT there would be many sets that would not satisfy condition 1 .

Here, we split the possible sets into cases:

$x=y=z$
Two of the variables are equal.
None of the variables are equal.

Case 1: There is only $1$ possible case: $x=y=z =10$ .

Case 2: Notice that the sets satisfying Case 2 can be split further into 6 types: (The subcases with an asterix satisfy condition 1 .)

$*x = y >z$ -------- $5$ ways when $z = 0,2,4,6,8$ or $<z$ ----------- $10$ ways when $z = 12, 14,16,18,20,22,24,26,28,30$
$y = z >x$ ------- $5$ ways or $*<x$ ------ $10$ ways
$x = z >y$ ------- $5$ ways or $<y$ ------- $10$ ways

As seen, only $2$ out of $6$ of the above subcases satisfy condition 1 . Hence, only $15$ ways worked out of $45$ ways.

Case 3: Now there are $496 -1 -45 = 450$ sets of $(x,y,z)$ that belong to Case 3. Notice that there are $3! = 6$ ways to permutate $x,y,z$ and out of the $6$ permutations, we only want $1$ ------ $x>y>z$ . Thus, we divide $450$ by $6 = 75.$

Hence, the number of sets that satisfies all conditions equal $1+15+75 = \boxed{91}$ .

Generalisation

Let the question be find the number of unordered nonnegative integer $(a,b,c)$ that satifies $3a +2b +c = n$ . Letting $n+2 \choose 2$ be $p$ , the answer is simply $1+ (\frac{n}{2}) + (\frac {p - 1 -\frac{3n}{2}}{6}) = 1 + \frac{n}{2} + \frac{n^{2}}{12}$ .

Happy Melodies - 7 years, 6 months ago

There is another approach, which doesn't require such extensive case analysis. This approach becomes extremely tedious when the points is 300, or 3000.

Hint: Apply Pick's Theorem.

Calvin Lin Staff - 7 years, 6 months ago

Here's an idea, which I think works. Basically, the number of 2 and 3 pointers determines the number of 1 pointers, so we can make the point (x,y) in the plane represent x three pointer and y two pointers. Then we have a triangle, and can use pick's theorem with some much easier counting of the border points to find the total number.

Henrik Boecken - 7 years, 6 months ago

@Henrik Boecken – Oh, I got it... Nice!

Jackal Jim - 7 years, 6 months ago

@Henrik Boecken – For the sake of completeness, here is how to do it:

We want to count the number of non-negative solutions to $a+2b+ 3c = 30$ . Another way to do so is to count the number of lattice points in the region $0 \leq x, 0 \leq y, 2x+3y \leq 30$ , and create the bijection $(a, b, c) \leftrightarrow (x, y)$ via $b=x, c=y, a= 30-2x-3y$ .

This region is a triangle with vertices $(0, 0), (0, 10), (15, 0)$ hence has area $\frac {1}{2} \times 10 \times 15 = 75$ . We apply Pick's Theorem. $L$ is the number of lattice points in this region, $B$ is the number of lattice points on the boundary, and $A$ is the area of this region, then $A = \frac {B}{2} + L - 1$ . We can calculate that $B = 10 + 15 + 5 = 30$ , so $L = 75 - 15 + 1 = 61$ . Hence, there are $61+30 =91$ possible ways.

Note: The bijection at the start reduces the problem from "3-D" to "2-D". However, we're actually already in 2-D, because $a + 2b + 3c = 30$ represents a 2-D plane in 3-D space. The bijection just reorientates your perspective to make it easier to work with.

Calvin Lin Staff - 2 years, 8 months ago

Pick's theorem states that the area of a polygon $P(A) = i$ , the number of interior lattice, $+\frac{b}{2}-1$ , where $b$ is the number of boundary lattice. I am not sure how to apply it to this question though...

Jackal Jim - 7 years, 6 months ago

@Jackal Jim – As you mentioned, Pick's Theorem tells us the number of lattice points in a given area. To apply it, we want to create a bijection between the set of basketball scores, and a certain set of lattice points in the plane.

Calvin Lin Staff - 7 years, 6 months ago

@Calvin Lin – Could you please post a solution applying Pick's Theorem? Thanks!

Neelansh Bute - 7 years, 6 months ago

@Neelansh Bute – (Idea by Henrik B. - comment above) I think this is what he meant:

Score! Pic. - if you cannot view it, go to this link Graph of triangle

As Henrik B. has mentioned earlier, let the $x$ -coordinates represent $x$ $3$ pointers and $y$ -coordinates represent $y$ $2$ pointers. The number of $2$ and $3$ pointers will determine the number of $1$ pointer - so we can conveniently leave the number of $1$ pointers out in our diagram.

Notice that since the total score is $30$ , the maximum values of $x$ and $y$ are $10$ and $15$ respectively. This is represented in the graph as the points $(10,0)$ and $(0,15)$ . Then, we join the two dots together to form the diagonal (red) of the triangle seen in the diagram. Next, we simply draw lines $x=0$ and $y=0$ to complete the triangle.

Pick's Theorem states that Area $= i + \frac{b}{2} -1$ , where $i$ represents the number of interior lattice points and $b$ represents the number of boundary points. We can count from the diagram that there are a total of $30$ boundary points on the triangle. The area of the triangle can be calculated easily - using the formula: Area of triangle $= \frac{1}{2} * base * height$ - $\frac{1}{2} * 10* 15 = 75$ .

Substituting the $2$ values we have obtained earlier into Pick's Theorem, we get $75 = i + \frac{30}{2} -1 = i + 15 -1$ .

From this, it is clear that $76 = i +15$ and $i = 76 - 15 = 61$ .

To find the number of $(x,y)$ that satisfies the question, it is simply the number of points within the boundary of the triangle $= i + b = 61 + 30 = \boxed{91}$ .

Happy Melodies - 7 years, 6 months ago

@Calvin Lin OMG!!! SIR!!! This is an AMAZING way to solve the question!!!! I mean how did Pick's theorem even strike you???? Like this is awesome!!!

Aaghaz Mahajan - 2 years, 8 months ago

@Aaghaz Mahajan –

I set the question based on wanting to use Pick's theorem. I tried to make it harder, but still wanted it to be brute-forcable, as most people would use brute force once they think they can.
It's not an uncommon technique for an Olympiad counting problem.

Calvin Lin Staff - 2 years, 8 months ago

@Calvin Lin – Well thanks anyways Sir!!!! Do u have any other amazing Olympiad tricks which you can suggest??

Aaghaz Mahajan - 2 years, 8 months ago

Nahom Yemane
Jan 2, 2014

Let $a$ denote the number of $1$ pointers, $b$ denote the number of $2$ pointers, $c$ denote the number of $3$ pointers.

So we have $a+2b+3c=30$ . Now just do a case bash

from $c=0$ giving $a+2b=30$ to $c=10$ giving $a+2b=0$

so first $c=0$ giving $a+2b=30$ $2b$ must even so $16$ different values for $2b$ and hence also for $a$ when $c=0$

Next $c=1$ giving $a+2b=27$ $2b$ must even so $14$ different values for $2b$ and hence also for $a$ when $c=1$

We can continue this getting a full list of:

$16 , 14 ,13, 11, 10, 8 , 7, 5,4, 2, 1$

$16 +14 +13 + 11+ 10 + 8 + 7+ 5 + 4 + 2 + 1=\boxed{91}$

Hafizh Ahsan Permana
Dec 11, 2013

The statement can be written as mathematical function, for examples $x+2y+3z=30$ , where $x$ is multiple of 1point , $y$ is multiple of 2 points, and $z$ is multiple of 3 points. Enter all possibilities that could happens for $x,y, and$ $z$ . Think a question in your mind like"what is the greatest points for z ?" it is 10. then "what is the minimum?"it is 0.

From the function subtitute your mind when $z=10$ then $x=0$ $and$ $y=0$ , next try $z=9$ $x=1$ $y=1$ and $z=9$ $x=3$ $y=0$ however what's on your mind that the sum is $30$ continue this pattern until the minimum point of $z$ where $z$ is $0$ and you'll see a pattern for sum of the possibilities.

$z=10(1 possibility), z=9(2 possibilities) ,z=8(4 possibilities), z=7(5 possibilities)$ continue this until $z=0(16 possibilities)$ and sum all the possibilities. Like this: $1 + 2 + 4 + 5 + 7 + 8 + 10 + 11 + 13 + 14 + 16=91$

or you can see this pattern more clearer $1 + (1+1) + (1+3) + (1+4) + (1+6) + (1+7) + (1+9) + (1+10) + (1+12) + (1+13) + (1+15)=91$ . You can try another ways, like points for x=30 and continue this until x=0 but i suggested z because it more simple(z is biggest multiply).

Aditya Joshi
Feb 6, 2014

First I'll write a possible configuration of number of throws that yields $30$ .

10 3's, 0 2's, 0 1's.

Let's write another one.

0 3's, 15 2's, 0 1's.

And another

0 3's, 15 2's, 30 1's

We notice that it is a combination of 1's 2's and 3's with some weight .

We can represent the number of throws required to get 30 as the equation $x + 2y + 3z = 30$ .

We need to find all integer solutions to $x,y,z.$

This is a linear diophantine and solving this ( look here for ways ) yields an a solution of $\boxed{91}$

Pebrudal Zanu
Dec 4, 2013

The problem equivalen with:

$x_1+x_2+...+x_n=30$

$n=10,11,12,...,30$

For

===============================

$n=10$ , The ways are $1$ way.

we choice $x_i=3$ for $10$

$i.e: (3,3,3,3,3,3,3,3,3,3)$

===============================

Only that the way,

===========================

$n=11$ , The ways are $2$ way

we choice $x_i=3$ for $8-9$

$i.e : (3,3,3,3,3,3,3,3,2,2,2)$ and $3,3,3,3,3,3,3,3,3,1,2)$

$3$ only can choice $8$ and $9$ ..

==========================

$n=12$ , The ways are $4$ way.

we choice $x_i=3$ for $6-9$

=============================

$n=13$ , The ways are $5$ way.

we choice $x_i=3$ for $4-8$

=============================

$n=14$ , The ways are $7$ way.

we choice $x_i=3$ for $2-8$

=============================

$n=15,16$ , The ways are $8$ way.

we choice $x_i=3$ for $0-7$

==============================

$n=17,18$ , The ways are $7$ way.

we choice $x_i=3$ for $0-6$

============================= $n=19,20$ , The ways are $6$ way.

we choice $x_i=3$ for $0-5$

==============================

$n=21,22$ , The ways are $5$ way.

we choice $x_i=3$ for $0-4$

===============================

$n=23,24$ , The ways are $4$ way.

we choice $x_i=3$ for $0-3$

================================

$n=25,26$ , The ways are $3$ way.

we choice $x_i=3$ for $0-2$

===============================

$n=27,28$ , The ways are $2$ way.

we choice $x_i=3$ for $0-1$

==============================

$n=29,30$ , The ways are $1$ way.

we choice $x_i=3$ for $0$

=============================

So, the answer are $1+2+4+5+7+2(1+2+3+4+5+6+7)=\fbox{91}$

I am sorry, forget :) $1+2+4+5+7+2(1+2+3+4+5+6+7+8$

pebrudal zanu - 7 years, 6 months ago

William Cui
Dec 3, 2013

If the player scores 10 3-pointers, there is only one possibility - that is already 30 points.

If the player scores 9 3's, he can score 0 or 1 2-pointer, with the rest being 1's, for a total of two possibilities.

If the player scores 8 3's, he can score 0, 1, 2, or 3 2-pointers, for a total of four possibilities.

If the player scores 7 3's, he can score 0, 1, 2, 3, or 4 2-pointers, for a total of five possibilities.

If the player scores 6 3's, he can score 0, 1, 2, 3, 4, 5, or 6 2-pointers, for a total of seven possibilities.

And so on. We can notice a pattern - the sequence is positive integers skipping every multiple of three. We want the first eleven terms of this (0-10 3-pointers), so we have

$1+2+4+5+7+8+10+11+13+14+16=\boxed{91}$

total possibilities, which is our desired answer. $\ \blacksquare$

Kshitij Khandelwal
Dec 3, 2013

You can approach this through common sense or through discrete math techniques.

Common sense approach: What's the greatest number of 3-point shots that might be included? 10. Way #1: 10 3pt shots. What if there was one less 3-point shot? That would be 27 points. There's only one way to make up the other 3 points. Way #2: 9 3pt shots, 1 2pt shot, 1 1pt shot. Way #3: 9 3pt shots, 3 1pt shots. And one less? That would be 24 points. But there are several ways of making the other 6 points, counting down from 3 2pt shots: Way #4: 8 3pt shots, 3 2pt shots. Way #5: 8 3pt shots, 2 2pt shots, 2 1pt shots. Way #6: 8 3pt shots, 1 2pt shot, 4 1pt shots. Way #7: 8 3pt shots, 6 1pt shots. Back to the 3pt shots. What about one less? That's 21 points, with 9 points from 1pt and 2pt shots. Way #8: 7 3pt shots, 4 2pt shots, 1 1pt shot. Way #9: 7 3pt shots, 3 2pt shots, 3 1pt shots. Way #10: 7 3pt shots, 2 2pt shots, 5 1pt shots. Way #11: 7 3pt shots, 1 2pt shot, 7 1pt shots. Way #12: 7 3pt shots, 9 1pt shots. And you continue on in that way until at the end, you are scoring all 30 points in 30 1pt shots.

Notice that once you've decided on the number of 3pt shots, and selected the maximum number of 2pt shots that can make up the difference, you can generate one more way by changing each of the 2pt shots to two 1pt shots, one at a time. So you'll have one more way for each 2pt shot you have in your first way.

I'll leave the specifics to you, but you end up with: 1 + (1+1) + (1+3) + (1+4) + (1+6) + (1+7) + (1+9) + (1+10) + (1+12) + (1+13) + (1+15) ways = 1 + 2 + 4 + 5 + 7 + 8 + 10 + 11 + 13 + 14 + 16 = 91 ways

The discrete math way is a little more difficult to understand without a background in it.

Can you provide more details about the discrete math way? How can we avoid doing such tedious case analysis?

Calvin Lin Staff - 7 years, 6 months ago

Arnav Shringi
Dec 2, 2013

Way #1: 10 3pt shots.

What if there was one less 3-point shot? That would be 27 points. There's only one way to make up the other 3 points.

Way #2: 9 3pt shots, 1 2pt shot, 1 1pt shot.

Way #3: 9 3pt shots, 3 1pt shots.

And one less? That would be 24 points. But there are several ways of making the other 6 points, counting down from 3 2pt shots:

Way #4: 8 3pt shots, 3 2pt shots.

Way #5: 8 3pt shots, 2 2pt shots, 2 1pt shots.

Way #6: 8 3pt shots, 1 2pt shot, 4 1pt shots.

Way #7: 8 3pt shots, 6 1pt shots.

Back to the 3pt shots. What about one less? That's 21 points, with 9 points from 1pt and 2pt shots.

Way #8: 7 3pt shots, 4 2pt shots, 1 1pt shot.

Way #9: 7 3pt shots, 3 2pt shots, 3 1pt shots.

Way #10: 7 3pt shots, 2 2pt shots, 5 1pt shots.

Way #11: 7 3pt shots, 1 2pt shot, 7 1pt shots.

Way #12: 7 3pt shots, 9 1pt shots.

And you continue on in that way until at the end, you are scoring all 30 points in 30 1pt shots.

I'll leave the specifics to you, but you end up with: 1 + (1+1) + (1+3) + (1+4) + (1+6) + (1+7) + (1+9) + (1+10) + (1+12) + (1+13) + (1+15) ways = 1 + 2 + 4 + 5 + 7 + 8 + 10 + 11 + 13 + 14 + 16 = * 91 ways*

Score!

The answer is 91.

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