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Comibatorics Question with Scenarios [#permalink]
 24 Mar 2010, 10:29
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A men's basket ball team assigns every player a two digit number for the back of his jersey. If the digit uses the digits 1 to 5, what is the maximum number of players that can join the league such that no player has a number with repeated digits (like 22) & no two players have the same number?? Guys, after answering the above question, kindly consider the following scenarios: Scenario 1: What will be the answer, if the repetition (eg 22) is allowed. Scenario 2: What will be the answer, if every player is assigned a "three" digit number from the digits 1 to 5 & the repetition of digits (eg 323 or 333) is not allowed. Scenario 3: What will be the answer, if every player is assigned a "three" digit number from the digits 1 to 5 & the repetition of digits (eg 323 or 333) is allowed. Well, I think if I can get some views on these scenarios, I can understand "Combinatorics". These scenarios emerged as a result of a discussion between myself & AtifS (an active member of the club & my partner). The source of question is MGMAT Guide while the 3 scenarios are our creation.
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Re: Comibatorics Question with Scenarios [#permalink]
24 Mar 2010, 10:46
Hussain15 wrote: A men's basket ball team assigns every player a two digit number for the back of his jersey. If the digit uses the digits 1 to 5, what is the maximum number of players that can join the league such that no player has a number with repeated digits (like 22) & no two players have the same number??
The source of question is MGMAT Guide while the 3 scenarios are our creation. 1st digit = 5 ways 2nd digit = 4 ways total = 5*4 = 20 players Please mention the OA
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Re: Comibatorics Question with Scenarios [#permalink]
24 Mar 2010, 10:49
Hussain15 wrote: A men's basket ball team assigns every player a two digit number for the back of his jersey. If the digit uses the digits 1 to 5, what is the maximum number of players that can join the league such that no player has a number with repeated digits (like 22) & no two players have the same number??
Guys, after answering the above question, kindly consider the following scenarios:
Scenario 1: What will be the answer, if the repetition (eg 22) is allowed.
Scenario 2: What will be the answer, if every player is assigned a "three" digit number from the digits 1 to 5 & the repetition of digits (eg 323 or 333) is not allowed.
Scenario 3: What will be the answer, if every player is assigned a "three" digit number from the digits 1 to 5 & the repetition of digits (eg 323 or 333) is allowed.
The source of question is MGMAT Guide while the 3 scenarios are our creation. Scenario1: if repetition allowed then its 5 * 5 = 25 players. Scenario 2: 5 * 4 * 3 = 60 players. Scenario 3: 5 * 5 * 5 = 125 players.
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VP
Status: The last round
Joined: 18 Jun 2009
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386
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Re: Comibatorics Question with Scenarios [#permalink]
24 Mar 2010, 12:14
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Re: Comibatorics Question with Scenarios [#permalink]
24 Mar 2010, 12:15
bangalorian2000 wrote: Hussain15 wrote: A men's basket ball team assigns every player a two digit number for the back of his jersey. If the digit uses the digits 1 to 5, what is the maximum number of players that can join the league such that no player has a number with repeated digits (like 22) & no two players have the same number??
Guys, after answering the above question, kindly consider the following scenarios:
Scenario 1: What will be the answer, if the repetition (eg 22) is allowed.
Scenario 2: What will be the answer, if every player is assigned a "three" digit number from the digits 1 to 5 & the repetition of digits (eg 323 or 333) is not allowed.
Scenario 3: What will be the answer, if every player is assigned a "three" digit number from the digits 1 to 5 & the repetition of digits (eg 323 or 333) is allowed.
The source of question is MGMAT Guide while the 3 scenarios are our creation. Scenario1: if repetition allowed then its 5 * 5 = 25 players. Scenario 2: 5 * 4 * 3 = 60 players. Scenario 3: 5 * 5 * 5 = 125 players. Nice! can you please! explain y did you take 5*5=5^2 for 1st scenario and y did you take 5*5*5=5^3 for 3rd scenario? I just wanted to know the logic behind t. Well I think 5 options are possible for 1st digit and similarly 5 for 2nd digit. Same goes for 3-digit number. But would like to know your explanation as there might be anything useful to know  Thanks, -A
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Re: Comibatorics Question with Scenarios [#permalink]
24 Mar 2010, 21:50
AtifS wrote: bangalorian2000 wrote: Hussain15 wrote: A men's basket ball team assigns every player a two digit number for the back of his jersey. If the digit uses the digits 1 to 5, what is the maximum number of players that can join the league such that no player has a number with repeated digits (like 22) & no two players have the same number??
Guys, after answering the above question, kindly consider the following scenarios:
Scenario 1: What will be the answer, if the repetition (eg 22) is allowed.
Scenario 2: What will be the answer, if every player is assigned a "three" digit number from the digits 1 to 5 & the repetition of digits (eg 323 or 333) is not allowed.
Scenario 3: What will be the answer, if every player is assigned a "three" digit number from the digits 1 to 5 & the repetition of digits (eg 323 or 333) is allowed.
The source of question is MGMAT Guide while the 3 scenarios are our creation. Scenario1: if repetition allowed then its 5 * 5 = 25 players. Scenario 2: 5 * 4 * 3 = 60 players. Scenario 3: 5 * 5 * 5 = 125 players. Nice! can you please! explain y did you take 5*5=5^2 for 1st scenario and y did you take 5*5*5=5^3 for 3rd scenario? I just wanted to know the logic behind t. Well I think 5 options are possible for 1st digit and similarly 5 for 2nd digit. Same goes for 3-digit number. But would like to know your explanation as there might be anything useful to know Thanks, -A The logic is same as you said, if digit repeat is allowed than ways to select 1st digit = 5 (either of 1,2,3,4,5) ways to select 2nd digit = 5 (either of 1,2,3,4,5) total = 5*5 = 25 for the case when number on player's shirt is of three digits then ways to select 1st digit = 5 (either of 1,2,3,4,5) ways to select 2nd digit = 5 (either of 1,2,3,4,5) ways to select 3rd digit = 5 (either of 1,2,3,4,5) total = 5*5*5 = 125
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Re: Comibatorics Question with Scenarios [#permalink]
24 Mar 2010, 23:22
@bangalorian! Now that's better even a newbie can understand that's why I asked you to explain it because others can understand it easily. P.S: Actually! I did have idea after discussion with Hussain15 but wasn't sure and also wanted others views.
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Re: Comibatorics Question with Scenarios
[#permalink]
24 Mar 2010, 23:22
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