A circular spinner is divided into 9 equal sectors numbered 1 through

Question

A circular spinner is divided into 9 equal sectors numbered 1 through 9. If the spinner is spun 3 times, how many different outcomes of the three spins result in a sum of 20?

A. 27
B. 36
C. 45
D. 72
E. 135

AdarshSambare · Answer

We need the number of ordered triples (a, b, c) such that:
a + b + c = 20
with each of a, b, c being an integer from 1 to 9.
Since the spins are separate, order matters.
Let
x = a − 1, y = b − 1, z = c − 1
Then x, y, z ≥ 0 and
x + y + z = 20 − 3 = 17
Also, since a, b, c ≤ 9,
x, y, z ≤ 8.
First count all nonnegative solutions to
x + y + z = 17.
Using stars and bars:
C(17 + 3 − 1, 3 − 1) = C(19, 2) = 171
Now subtract solutions where at least one variable exceeds 8.
Let x ≥ 9.
Set x' = x − 9 ≥ 0.
Then
x' + y + z = 8
Number of solutions:
C(10, 2) = 45
Similarly for y ≥ 9 and z ≥ 9, giving
3 × 45 = 135
Now add back intersections.
If x ≥ 9 and y ≥ 9, then
x' + y' + z = -1
impossible.
Likewise all pairwise and triple intersections are impossible.
Therefore:
Total valid solutions
= 171 − 135
= 36
Answer: Option: B 36

naveeng15 · Answer

is there any easy way to do these kind of problems ?  KarishmaB

Daphnee · Answer

OA:
This is the fastest and most straightforward way I could think of but please let me know if there is a better way :v

First, determine the possible combinations of three numbers that sum to 20, using an decreasing order (x≥y≥z) to avoid confusion or duplication. Then, calculate the number of permutations for each combination.
Combination (9, 9, 2) → 3 permutations (3!/2!). 
Combination (9, 8, 3) → 6 permutations (3!). 
Combination (9, 7, 4) → 6 permutations (3!). 
Combination (9, 6, 5) → 6 permutations (3!). 
Combination (8, 8, 4) → 3 permutations (3!/2!). 
Combination (8, 7, 5) → 6 permutations (3!). 
Combination (8, 6, 6) → 3 permutations (3!/2!). 
Combination (7, 7, 6) → 3 permutations (3!/2!). 
Total outcomes: 36 ways.

egmat · Answer

Hi naveeng15,

Yes, there's a much lighter way, and you don't need stars-and-bars or the inclusion-exclusion subtraction at all.

The trick is  symmetry . Each sector is numbered  1  to  9 , and 1 and 9 are mirror images around the center value 5 (so are 2 and 8, 3 and 7, and so on). In general, replacing every value  v  with  10 - v  is a perfect one-to-one relabeling of the spinner.

Watch what that does to a sum. If three spins add to  20 , then after the swap they add to:

(10 - a) + (10 - b) + (10 - c) = 30 - 20 =  10

So every triple that sums to  20  pairs up with exactly one triple that sums to  10 . That means:

(number of ways to get 20) = (number of ways to get 10)

And counting ways to get  10  is far friendlier, because the numbers are small and you never bump into the "max is 9" ceiling.

Count the easy side

For an ordered triple (a, b, c) with each from 1 to 9 summing to  10 , shift each down by 1 (let a' = a - 1, etc.). Now a' + b' + c' =  7 , each a' from 0 to 8. Since 7 is already less than 9, the upper limit never bites - no cases to subtract.

Ordered non-negative solutions of a' + b' + c' = 7 is a clean stars-and-bars: C(7 + 2, 2) = C(9, 2) =  36 .

So the answer is  36 .

The takeaway:  whenever a "high sum" count looks painful, check whether reflecting the values (here v -> 10 - v) turns it into a small "low sum" count. Same answer, a fraction of the work.

Answer: B

A circular spinner is divided into 9 equal sectors numbered 1 through

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