In a 4 person race, medals are awarded to the fastest 3 sprinters. The first-place sprinter receives a gold medal, the second-place sprinter receives a silver medal, and the third-place sprinter receives a bronze medal. In the event of a tie, the tied sprinters receive the same color medal. (For example, if there is a two-way tie for first-place, the top two sprinters receive gold medals, the next-fastest sprinter receives a silver medal, and no bronze medal is awarded). Assuming that exactly three medals are awarded, and that the three medal winners stand together with their medals to form a Winner's triangle, how many different Winners' triangles are possible?
In a 4 person race, medals are awarded to the fastest 3 sprinters. The first-place sprinter receives a gold medal, the second-place sprinter receives a silver medal, and the third-place sprinter receives a bronze medal. In the event of a tie, the tied sprinters receive the same color medal. (For example, if there is a two-way tie for first-place, the top two sprinters receive gold medals, the next-fastest sprinter receives a silver medal, and no bronze medal is awarded). Assuming that exactly three medals are awarded, and that the three medal winners stand together with their medals to form a Winner's triangle, how many different Winners' triangles are possible?
The correct option is B 52
First, let's consider the different medal combinations that can be awarded to the 3 winners:
(1) If there are NO TIES then the three medals awarded are: GOLD, SILVER, BRONZE.
(2) What if there is a 2-WAY tie?
--If there is a 2-WAY tie for FIRST, then the medals awarded are: GOLD, GOLD, SILVER.
--If there is a 2-WAY tie for SECOND, then the medals awarded are: GOLD, SILVER, SILVER.
--There cannot be a 2-WAY tie for THIRD (because exactly three medals are awarded in total).
(3) What if there is a 3-WAY tie?
--If there is a 3-WAY tie for FIRST, then the medals awarded are: GOLD, GOLD, GOLD.
--There are no other possible 3-WAY ties.
Thus, there are 4 possible medal combinations:
(1) G, S, B (2) G, G, S (3) G, S, S (4) G, G, G
COMBINATION 1: Gold, Silver, Bronze
Number of possibilities = 4×3×2 =24 different Winner's triangles that will contain 1 GOLD, 1 SILVER, and 1 BRONZE medalist.
COMBINATION 2: Gold, Gold, Silver.
Selection =4C3 and arrangement =4!3!(4−3)!= 4×3=4 ways
COMBINATION 3: Gold, Silver, Silver.
Using the same reasoning as for Combination 2, we see that there are 24 possible Winner's triangles, but only 12 unique Winner's triangles that contain 1 GOLD medalist and 2 SILVER medalists.
COMBINATION 4: Gold, Gold, Gold
Finally , then we have the following:
Selection =4C3 and arrangement =3!2!= 4 ×1=4 ways
Thus, there are 24 + 12 + 12 + 4 = 52 unique Winner's triangles.
The correct answer is B.
52
First, let's consider the different medal combinations that can be awarded to the 3 winners:
(1) If there are NO TIES then the three medals awarded are: GOLD, SILVER, BRONZE.
(2) What if there is a 2-WAY tie?
--If there is a 2-WAY tie for FIRST, then the medals awarded are: GOLD, GOLD, SILVER.
--If there is a 2-WAY tie for SECOND, then the medals awarded are: GOLD, SILVER, SILVER.
--There cannot be a 2-WAY tie for THIRD (because exactly three medals are awarded in total).
(3) What if there is a 3-WAY tie?
--If there is a 3-WAY tie for FIRST, then the medals awarded are: GOLD, GOLD, GOLD.
--There are no other possible 3-WAY ties.
Thus, there are 4 possible medal combinations:
(1) G, S, B (2) G, G, S (3) G, S, S (4) G, G, G
COMBINATION 1: Gold, Silver, Bronze
Number of possibilities = 4×3×2 =24 different Winner's triangles that will contain 1 GOLD, 1 SILVER, and 1 BRONZE medalist.
COMBINATION 2: Gold, Gold, Silver.
Selection =4C3 and arrangement =4!3!(4−3)!= 4×3=4 ways
COMBINATION 3: Gold, Silver, Silver.
Using the same reasoning as for Combination 2, we see that there are 24 possible Winner's triangles, but only 12 unique Winner's triangles that contain 1 GOLD medalist and 2 SILVER medalists.
COMBINATION 4: Gold, Gold, Gold
Finally , then we have the following:
Selection =4C3 and arrangement =3!2!= 4 ×1=4 ways
Thus, there are 24 + 12 + 12 + 4 = 52 unique Winner's triangles.
The correct answer is B.
