Permutation and combination(ALREADY SOLVED) [closed]

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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Find the number of teams of 11 that can be selected from a group of 15 players if the youngest 2 players and at most one of the oldest two players are to be included



Ans:385




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closed as off-topic by Tanner Swett, gammatester, Shaun, John Douma, Xander Henderson Aug 27 at 0:14


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Tanner Swett, gammatester, Shaun, John Douma, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.








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    Welcome to Stack Exchange! Posts which contain nothing more than a problem statement (such as your post) are unlikely to receive answers. I recommend giving us some additional context (what you understand about the problem, what you've tried so far, and so forth) in order to help us guide you to the appropriate help. See "How to ask a good question" for more useful information.
    – Tanner Swett
    Aug 26 at 21:50














up vote
-4
down vote

favorite
1












Find the number of teams of 11 that can be selected from a group of 15 players if the youngest 2 players and at most one of the oldest two players are to be included



Ans:385




statistics




share|cite|improve this question














closed as off-topic by Tanner Swett, gammatester, Shaun, John Douma, Xander Henderson Aug 27 at 0:14


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Tanner Swett, gammatester, Shaun, John Douma, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 2




    Welcome to Stack Exchange! Posts which contain nothing more than a problem statement (such as your post) are unlikely to receive answers. I recommend giving us some additional context (what you understand about the problem, what you've tried so far, and so forth) in order to help us guide you to the appropriate help. See "How to ask a good question" for more useful information.
    – Tanner Swett
    Aug 26 at 21:50












up vote
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up vote
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down vote

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1





Find the number of teams of 11 that can be selected from a group of 15 players if the youngest 2 players and at most one of the oldest two players are to be included



Ans:385




statistics




share|cite|improve this question














Find the number of teams of 11 that can be selected from a group of 15 players if the youngest 2 players and at most one of the oldest two players are to be included



Ans:385





statistics





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 27 at 3:34

























asked Aug 26 at 21:32









Law

11




11




closed as off-topic by Tanner Swett, gammatester, Shaun, John Douma, Xander Henderson Aug 27 at 0:14


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Tanner Swett, gammatester, Shaun, John Douma, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Tanner Swett, gammatester, Shaun, John Douma, Xander Henderson Aug 27 at 0:14


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Tanner Swett, gammatester, Shaun, John Douma, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 2




    Welcome to Stack Exchange! Posts which contain nothing more than a problem statement (such as your post) are unlikely to receive answers. I recommend giving us some additional context (what you understand about the problem, what you've tried so far, and so forth) in order to help us guide you to the appropriate help. See "How to ask a good question" for more useful information.
    – Tanner Swett
    Aug 26 at 21:50












  • 2




    Welcome to Stack Exchange! Posts which contain nothing more than a problem statement (such as your post) are unlikely to receive answers. I recommend giving us some additional context (what you understand about the problem, what you've tried so far, and so forth) in order to help us guide you to the appropriate help. See "How to ask a good question" for more useful information.
    – Tanner Swett
    Aug 26 at 21:50







2




2




Welcome to Stack Exchange! Posts which contain nothing more than a problem statement (such as your post) are unlikely to receive answers. I recommend giving us some additional context (what you understand about the problem, what you've tried so far, and so forth) in order to help us guide you to the appropriate help. See "How to ask a good question" for more useful information.
– Tanner Swett
Aug 26 at 21:50




Welcome to Stack Exchange! Posts which contain nothing more than a problem statement (such as your post) are unlikely to receive answers. I recommend giving us some additional context (what you understand about the problem, what you've tried so far, and so forth) in order to help us guide you to the appropriate help. See "How to ask a good question" for more useful information.
– Tanner Swett
Aug 26 at 21:50










1 Answer
1






active

oldest

votes

















up vote
1
down vote













If I understand the question correctly...



You can partition your "set" of players into the different groups (youngest, "middle", oldest) which simplifies your problem. We have the...



2 youngest players



11 "middle aged" players



2 oldest players



Now you want 2 out of the 2 youngest players, 1 of the 2 oldest players, and thus 8 out of the 11 "other" players OR 2 out of the 2 youngest players, 0 of the 2 oldest players, and thus 9 out of the 11 "other" players.



The number of possible teams is then: $binom22cdotbinom21cdotbinom118+binom22cdotbinom20cdotbinom119 = 1cdot2cdot165 +1cdot1cdot55=385$






share|cite|improve this answer






















  • Thats not the answer but nice try
    – Law
    Aug 27 at 3:02










  • I've got it by using your approach. You have not taken into consideration the at most part which means 1 or 0 of the oldest two player must be included. The answer is (2c2*11c9)+(2c2*2c1*11c8)
    – Law
    Aug 27 at 3:09










  • ah, sorry, "at most" one of the oldest players, overread that, sry, edited the answer
    – BruceBrain
    Aug 27 at 8:40


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote













If I understand the question correctly...



You can partition your "set" of players into the different groups (youngest, "middle", oldest) which simplifies your problem. We have the...



2 youngest players



11 "middle aged" players



2 oldest players



Now you want 2 out of the 2 youngest players, 1 of the 2 oldest players, and thus 8 out of the 11 "other" players OR 2 out of the 2 youngest players, 0 of the 2 oldest players, and thus 9 out of the 11 "other" players.



The number of possible teams is then: $binom22cdotbinom21cdotbinom118+binom22cdotbinom20cdotbinom119 = 1cdot2cdot165 +1cdot1cdot55=385$






share|cite|improve this answer






















  • Thats not the answer but nice try
    – Law
    Aug 27 at 3:02










  • I've got it by using your approach. You have not taken into consideration the at most part which means 1 or 0 of the oldest two player must be included. The answer is (2c2*11c9)+(2c2*2c1*11c8)
    – Law
    Aug 27 at 3:09










  • ah, sorry, "at most" one of the oldest players, overread that, sry, edited the answer
    – BruceBrain
    Aug 27 at 8:40















up vote
1
down vote













If I understand the question correctly...



You can partition your "set" of players into the different groups (youngest, "middle", oldest) which simplifies your problem. We have the...



2 youngest players



11 "middle aged" players



2 oldest players



Now you want 2 out of the 2 youngest players, 1 of the 2 oldest players, and thus 8 out of the 11 "other" players OR 2 out of the 2 youngest players, 0 of the 2 oldest players, and thus 9 out of the 11 "other" players.



The number of possible teams is then: $binom22cdotbinom21cdotbinom118+binom22cdotbinom20cdotbinom119 = 1cdot2cdot165 +1cdot1cdot55=385$






share|cite|improve this answer






















  • Thats not the answer but nice try
    – Law
    Aug 27 at 3:02










  • I've got it by using your approach. You have not taken into consideration the at most part which means 1 or 0 of the oldest two player must be included. The answer is (2c2*11c9)+(2c2*2c1*11c8)
    – Law
    Aug 27 at 3:09










  • ah, sorry, "at most" one of the oldest players, overread that, sry, edited the answer
    – BruceBrain
    Aug 27 at 8:40













up vote
1
down vote










up vote
1
down vote









If I understand the question correctly...



You can partition your "set" of players into the different groups (youngest, "middle", oldest) which simplifies your problem. We have the...



2 youngest players



11 "middle aged" players



2 oldest players



Now you want 2 out of the 2 youngest players, 1 of the 2 oldest players, and thus 8 out of the 11 "other" players OR 2 out of the 2 youngest players, 0 of the 2 oldest players, and thus 9 out of the 11 "other" players.



The number of possible teams is then: $binom22cdotbinom21cdotbinom118+binom22cdotbinom20cdotbinom119 = 1cdot2cdot165 +1cdot1cdot55=385$






share|cite|improve this answer














If I understand the question correctly...



You can partition your "set" of players into the different groups (youngest, "middle", oldest) which simplifies your problem. We have the...



2 youngest players



11 "middle aged" players



2 oldest players



Now you want 2 out of the 2 youngest players, 1 of the 2 oldest players, and thus 8 out of the 11 "other" players OR 2 out of the 2 youngest players, 0 of the 2 oldest players, and thus 9 out of the 11 "other" players.



The number of possible teams is then: $binom22cdotbinom21cdotbinom118+binom22cdotbinom20cdotbinom119 = 1cdot2cdot165 +1cdot1cdot55=385$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Aug 27 at 8:42

























answered Aug 26 at 21:56









BruceBrain

114




114











  • Thats not the answer but nice try
    – Law
    Aug 27 at 3:02










  • I've got it by using your approach. You have not taken into consideration the at most part which means 1 or 0 of the oldest two player must be included. The answer is (2c2*11c9)+(2c2*2c1*11c8)
    – Law
    Aug 27 at 3:09










  • ah, sorry, "at most" one of the oldest players, overread that, sry, edited the answer
    – BruceBrain
    Aug 27 at 8:40

















  • Thats not the answer but nice try
    – Law
    Aug 27 at 3:02










  • I've got it by using your approach. You have not taken into consideration the at most part which means 1 or 0 of the oldest two player must be included. The answer is (2c2*11c9)+(2c2*2c1*11c8)
    – Law
    Aug 27 at 3:09










  • ah, sorry, "at most" one of the oldest players, overread that, sry, edited the answer
    – BruceBrain
    Aug 27 at 8:40
















Thats not the answer but nice try
– Law
Aug 27 at 3:02




Thats not the answer but nice try
– Law
Aug 27 at 3:02












I've got it by using your approach. You have not taken into consideration the at most part which means 1 or 0 of the oldest two player must be included. The answer is (2c2*11c9)+(2c2*2c1*11c8)
– Law
Aug 27 at 3:09




I've got it by using your approach. You have not taken into consideration the at most part which means 1 or 0 of the oldest two player must be included. The answer is (2c2*11c9)+(2c2*2c1*11c8)
– Law
Aug 27 at 3:09












ah, sorry, "at most" one of the oldest players, overread that, sry, edited the answer
– BruceBrain
Aug 27 at 8:40





ah, sorry, "at most" one of the oldest players, overread that, sry, edited the answer
– BruceBrain
Aug 27 at 8:40



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