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13 Jul 2018, 05:28
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In a group of 10 people, 3 prizes od $20, $10, and $5 are to be given. How many ways can the prizes be distributed?
answer is 720
the yt video says that orders matters in this....but how and why?
answer is 720
the yt video says that orders matters in this....but how and why?
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
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13 Jul 2018, 07:59
Kudos
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Lets think of the above question like this.
In a group of 10 people, 3 prizes of $20, $10, and $5 are to be given.
Lets assume there is another prize(Loser) that is given to all the other people who are not first, second or third.
Now the question becomes, in how many ways the four prizes can be given to 10 people
=> We need to pick 3 people for first three prizes and the remaining seven people we don't need to pick. Once we pick 1, 2 and 3 the remaining 7 will automatically gets the Loser prize.
Total number of ways to pick 10 people = \(10!\)
Since the above number also includes different ways of picking the remaining 7 people and we actually don't need to pick remaining 7, we divide 10! by 7!
=> \(\frac{10!}{7!}\) = 720
Hope this helps your understanding.
In a group of 10 people, 3 prizes of $20, $10, and $5 are to be given.
Lets assume there is another prize(Loser) that is given to all the other people who are not first, second or third.
Now the question becomes, in how many ways the four prizes can be given to 10 people
=> We need to pick 3 people for first three prizes and the remaining seven people we don't need to pick. Once we pick 1, 2 and 3 the remaining 7 will automatically gets the Loser prize.
Total number of ways to pick 10 people = \(10!\)
Since the above number also includes different ways of picking the remaining 7 people and we actually don't need to pick remaining 7, we divide 10! by 7!
=> \(\frac{10!}{7!}\) = 720
Hope this helps your understanding.
13 Jul 2018, 08:48
Kudos
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Also check this thread permutations-combinations-help-is-on-the-way-10838.html
It has different types of permutation and combination problems to practice on.
It has different types of permutation and combination problems to practice on.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
