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Out of the 7 participants, the first winner who receives Platinum can be chosen in 7 ways

Out of the remaining 6 participants, the second winner who receives Gold can be chosen in 6 ways.

Out of the remaining 5 participants, the next 2 winners who receive Silver can be chosen in 5C2 = 10 ways.

Now we don't need to choose anything because the remaining 3 participants will automatically receive the Bronze medal.

Hence, total number of ways = 7*6*10 = 420

Only option choice B equates to 420 as 7!/(2!3!) = 5040/2*6 = 420

Hence, B.

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Seven people enter a race. There are 4 types of medals. Winner - platinum, runner up - gold, next two racers - silver and the last 3 - bronze. Ways to do it?

A. \(\frac{7!}{3!}\)

B. \(\frac{7!}{2!3!}\)

C. \(1*1*2!*3!\)

D. \(7*6*2!*3!\)

E. \(7*6*5*4\)

MartyTargetTestPrep

Happy Wednesday!
I tried solving this using the combination and then the permutation formula instead of the anagram method, treating each type of metal separately and was not able to match the correct answer.

I first did 7!/6! + 7!/6! + 7!/2!5! +7!/3!4! which was too small
Then I did
7!/6! + 6!/5! +5!/3! + 3!/0 (so this could not be correct due to the 0 in the denominator).

I would be so appreciative for your input.
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woohoo921
Bunuel
Seven people enter a race. There are 4 types of medals. Winner - platinum, runner up - gold, next two racers - silver and the last 3 - bronze. Ways to do it?

A. \(\frac{7!}{3!}\)

B. \(\frac{7!}{2!3!}\)

C. \(1*1*2!*3!\)

D. \(7*6*2!*3!\)

E. \(7*6*5*4\)

MartyTargetTestPrep

Happy Wednesday!
I tried solving this using the combination and then the permutation formula instead of the anagram method, treating each type of metal separately and was not able to match the correct answer.

I first did 7!/6! + 7!/6! + 7!/2!5! +7!/3!4! which was too small
Then I did
7!/6! + 6!/5! +5!/3! + 3!/0 (so this could not be correct due to the 0 in the denominator).

I would be so appreciative for your input.
You can't add the people who get the medals. The outcomes hinge each other. Try using the slots method instead.
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woohoo921
Bunuel
Seven people enter a race. There are 4 types of medals. Winner - platinum, runner up - gold, next two racers - silver and the last 3 - bronze. Ways to do it?

A. \(\frac{7!}{3!}\)

B. \(\frac{7!}{2!3!}\)

C. \(1*1*2!*3!\)

D. \(7*6*2!*3!\)

E. \(7*6*5*4\)

MartyTargetTestPrep

Happy Wednesday!
I tried solving this using the combination and then the permutation formula instead of the anagram method, treating each type of metal separately and was not able to match the correct answer.

I first did 7!/6! + 7!/6! + 7!/2!5! +7!/3!4! which was too small
Then I did
7!/6! + 6!/5! +5!/3! + 3!/0 (so this could not be correct due to the 0 in the denominator).

I would be so appreciative for your input.
You can't add the people who get the medals. The outcomes hinge each other. Try using the slots method instead.

Thank you for your response. Is the slot method = anagram method?
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Try using the slots method instead.

Thank you for your response. Is the slot method = anagram method?
The slots method is the method that involves creating a slot for each thing you have to choose.

So, for instance, in this case, the first slot would be for the person getting the platinum medal, the second slot would be for the person getting the gold medal, etc. You could also experiment with not using any slot for bronze winners since the bronze winners are just the ones who didn't get any other medal.

I like the slots method because it paints a picture of what's going on.

Look it up. You'll get a much clearer picture of how to answer these questions.
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woohoo921
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Try using the slots method instead.

Thank you for your response. Is the slot method = anagram method?
The slots method is the method that involves creating a slot for each thing you have to choose.

So, for instance, in this case, the first slot would be for the person getting the platinum medal, the second slot would be for the person getting the gold medal, etc. You could also experiment with not using any slot for bronze winners since the bronze winners are just the ones who didn't get any other medal.

I like the slots method because it paints a picture of what's going on.

Look it up. You'll get a much clearer picture of how to answer these questions.

MartyTargetTestPrep
Thank you, Marty! To clarify, the wording of the problem seemed similar to this one: https://gmatclub.com/forum/a-coach-is-t ... 00211.html

So, that is why I was trying to use combination formulas. What is the difference between that problem and this problem that prevents you from using a series of combination formulas to solve?

Thank you again.
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MartyTargetTestPrep
The slots method is the method that involves creating a slot for each thing you have to choose.

So, for instance, in this case, the first slot would be for the person getting the platinum medal, the second slot would be for the person getting the gold medal, etc. You could also experiment with not using any slot for bronze winners since the bronze winners are just the ones who didn't get any other medal.

I like the slots method because it paints a picture of what's going on.

Look it up. You'll get a much clearer picture of how to answer these questions.

MartyTargetTestPrep
Thank you, Marty! To clarify, the wording of the problem seemed similar to this one: https://gmatclub.com/forum/a-coach-is-t ... 00211.html

So, that is why I was trying to use combination formulas. What is the difference between that problem and this problem that prevents you from using a series of combination formulas to solve?

Thank you again.
Not much, but you didn't do the same thing.

Did you use addition to answer that other one?

You have to think about the logic of what you're doing.

You're having trouble with these questions because you're kind of just throwing math at them rather than analyzing the scenario and then deciding what you need to do to logically present the scenario in mathematical terms.
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