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How many 3 digit numbers can we make such that two of the digits are [#permalink]
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07 Jul 2018, 04:28
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How many 3 digit numbers can we make such that two of the digits are same and non of the digits equal zero? A. 60 B. 72 C. 150 D. 216 E. 280
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Re: How many 3 digit numbers can we make such that two of the digits are [#permalink]
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07 Jul 2018, 05:42
Following are the possibilities 
**# or *## or *#* 9*8 or 9*8 or 9*8 ways i.e., 72*3 = 216 ways. Thus, D is the answer.
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Re: How many 3 digit numbers can we make such that two of the digits are [#permalink]
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07 Jul 2018, 10:50
We have to consider several scenarios 1) xxy 9*1*8=72 2) xyx 9*8*1=72 3) yxx 8*9*1=72
3*72=216 Where x is the same digits and y is the different one
In My Opinion Ans: D



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Re: How many 3 digit numbers can we make such that two of the digits are [#permalink]
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07 Jul 2018, 12:44
Since the digits are non zero, hence 9 digits {1,2,....9} available to chose from
First digit = 9C1 = 9 Second digit = same as first digit = 1 Third digit = 8C1 = 8
Arrangements possible = 3!/2! = 3
Hence total 3 digit such #'s possible = 9*1*8*3 = 216
Answer D.
Thanks, GyM



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Re: How many 3 digit numbers can we make such that two of the digits are [#permalink]
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07 Jul 2018, 13:13
GyMrAT wrote: Since the digits are non zero, hence 9 digits {1,2,....9} available to chose from
First digit = 9C1 = 9 Second digit = same as first digit = 1 Third digit = 8C1 = 8
Arrangements possible = 3!/2! = 3
Hence total 3 digit such #'s possible = 9*1*8*3 = 216
Answer D. Thanks, GyM Thanks everyone for the detailed answer however I have the following few questions. 1. Why use 9C1 and 8C1? Usually when we use Fundamental Counting Principle, we simply put the # of possibilities at each step. If we didn't have the restriction that two digits are the same then we would have solved it by 9*8*7. Is combination generally hidden in FCP? 2. Also why did you say "Arrangements Possible = 3!/2!"? I thought the permutation took care of all arrangements when we compute 9*1*8 (XXY)  as the permutation definition says that if an event can occur in m ways and a second event in n ways then the two events can occur in m*n ways. When we have the general form XYZ, is 3!/3! hidden in the calculation? Regards,



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Re: How many 3 digit numbers can we make such that two of the digits are [#permalink]
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09 Jul 2018, 23:11
computerbot wrote: Thanks everyone for the detailed answer however I have the following few questions.
1. Why use 9C1 and 8C1? Usually when we use Fundamental Counting Principle, we simply put the # of possibilities at each step. If we didn't have the restriction that two digits are the same then we would have solved it by 9*8*7. Is combination generally hidden in FCP?
Since we are choosing one digit at a time to fill a slot. Suppose you were given 3 different bowls & 9 different marbles and told that each bowl should have 2 marbles. You will do that in 9C2*7C2*5C2
2. Also why did you say "Arrangements Possible = 3!/2!"? I thought the permutation took care of all arrangements when we compute 9*1*8 (XXY)  as the permutation definition says that if an event can occur in m ways and a second event in n ways then the two events can occur in m*n ways. When we have the general form XYZ, is 3!/3! hidden in the calculation?
The keyword you are missing is "Independent" events.
Since all the slots (hundred's place, tens place & units place) are distinct, the repeated digit has three options to be in. When we have XXY arrangement, it means that we are selecting one place (either hundred's or tens place) for the repeated digit & arranging the other two places with 2 unique digits.
Consider Hundreds place for repeated digit is selected, we have 9 options for the tens Place, 8 options for the units place & 1 option for the hundred's place
For e.g. We select 1, we have arrangements,
When 1 is in hundred's place as 112, 113, 114, 115,.....119, 121, 131, 141,.....191  16 #'s
Now 1 in tens place, we have, 112, 113, 114,.....119, 911, 811, 711,....211  16#'s
Now 1 in units place, we have, 121, 131, 141,....191, 911, 811, 711,......211  16#'s
So we have Total Arrangements = 16 + 16 + 16 = 48, but as you can see half of which are duplicates. Hence we consider only 24 unique ones.
You can perform this for all the other remaining 8 digits, getting only 24 unique arrangements.
If you want to master Combinatorics, i suggest, if you have not already, then devote an obscene amount of time in assimilating the concepts explained by Karishma in the below article. Its a Combinatorics Bible. https://gmatclub.com/forum/combinatoric ... l#p1579442Hope that helps. Thanks, GyM



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Re: How many 3 digit numbers can we make such that two of the digits are [#permalink]
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10 Jul 2018, 00:23
computerbot wrote: GyMrAT wrote: Since the digits are non zero, hence 9 digits {1,2,....9} available to chose from
First digit = 9C1 = 9 Second digit = same as first digit = 1 Third digit = 8C1 = 8
Arrangements possible = 3!/2! = 3
Hence total 3 digit such #'s possible = 9*1*8*3 = 216
Answer D. Thanks, GyM Thanks everyone for the detailed answer however I have the following few questions. 1. Why use 9C1 and 8C1? Usually when we use Fundamental Counting Principle, we simply put the # of possibilities at each step. If we didn't have the restriction that two digits are the same then we would have solved it by 9*8*7. Is combination generally hidden in FCP? 2. Also why did you say "Arrangements Possible = 3!/2!"? I thought the permutation took care of all arrangements when we compute 9*1*8 (XXY)  as the permutation definition says that if an event can occur in m ways and a second event in n ways then the two events can occur in m*n ways. When we have the general form XYZ, is 3!/3! hidden in the calculation? Regards, Hi mate, U are getting too technical For a 3 digit number, wherein u cannot have a zero and just 2 numbers repeated, make a slot  _ _ _ In first slot you can have any number from 1 to 9, so you have 9 options there. Second slot (the number being same as slot 1), you have only 1 option. Third slot can have any number other than the one in slot 1 and 2, giving you only 8 options. So you get 9*1*8 = 72 Note, you can have these numbers in 3 orders. Hence multiply 72 by 3. Did that help? Is it any easier for you now?



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Re: How many 3 digit numbers can we make such that two of the digits are [#permalink]
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12 Jul 2018, 03:22
GyMrAT wrote: If you want to master Combinatorics, i suggest, if you have not already, then devote an obscene amount of time in assimilating the concepts explained by Karishma in the below article. Its a Combinatorics Bible. https://gmatclub.com/forum/combinatoric ... l#p1579442Hope that helps. Thanks, GyM Thanks. I haven't gone through it yet but I needed it. Blueblu wrote: Hi mate, U are getting too technical For a 3 digit number, wherein u cannot have a zero and just 2 numbers repeated, make a slot  _ _ _ In first slot you can have any number from 1 to 9, so you have 9 options there. Second slot (the number being same as slot 1), you have only 1 option. Third slot can have any number other than the one in slot 1 and 2, giving you only 8 options. So you get 9*1*8 = 72 Note, you can have these numbers in 3 orders. Hence multiply 72 by 3. Did that help? Is it any easier for you now? Blueblu... thanks for the response. What I don't understand is that if we had all three distinct digits XYZ, we would have simply multiplied by 9*8*7. But XYZ can also be arranged in the following 6 ways but we don't multiple X*Y*Z with 6. XYZ XZY YXZ YZX ZXY ZYX Why is that so????



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Re: How many 3 digit numbers can we make such that two of the digits are [#permalink]
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13 Jul 2018, 01:36
computerbot wrote: GyMrAT wrote: If you want to master Combinatorics, i suggest, if you have not already, then devote an obscene amount of time in assimilating the concepts explained by Karishma in the below article. Its a Combinatorics Bible. https://gmatclub.com/forum/combinatoric ... l#p1579442Hope that helps. Thanks, GyM Thanks. I haven't gone through it yet but I needed it. Blueblu wrote: Hi mate, U are getting too technical For a 3 digit number, wherein u cannot have a zero and just 2 numbers repeated, make a slot  _ _ _ In first slot you can have any number from 1 to 9, so you have 9 options there. Second slot (the number being same as slot 1), you have only 1 option. Third slot can have any number other than the one in slot 1 and 2, giving you only 8 options. So you get 9*1*8 = 72 Note, you can have these numbers in 3 orders. Hence multiply 72 by 3. Did that help? Is it any easier for you now? Blueblu... thanks for the response. What I don't understand is that if we had all three distinct digits XYZ, we would have simply multiplied by 9*8*7. But XYZ can also be arranged in the following 6 ways but we don't multiple X*Y*Z with 6. XYZ XZY YXZ YZX ZXY ZYX Why is that so???? You are confusing distinct digits with the number of options. All of your digits X Y and Z are one of the digits from 1 to 9. Say you pick 678 and compare it with 876; in the first slot you put 9 which takes care of all 9 options including the digits 6,7 and 8 and so forth. This is why you don't multiply 9*8*7 with 3. I agree with GymArt  read Karishma's bible to get a clarity of your concepts.



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Re: How many 3 digit numbers can we make such that two of the digits are [#permalink]
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14 Jul 2018, 12:27
Blueblu wrote: computerbot wrote: GyMrAT wrote: If you want to master Combinatorics, i suggest, if you have not already, then devote an obscene amount of time in assimilating the concepts explained by Karishma in the below article. Its a Combinatorics Bible. https://gmatclub.com/forum/combinatoric ... l#p1579442Hope that helps. Thanks, GyM Thanks. I haven't gone through it yet but I needed it. Blueblu wrote: Hi mate, U are getting too technical For a 3 digit number, wherein u cannot have a zero and just 2 numbers repeated, make a slot  _ _ _ In first slot you can have any number from 1 to 9, so you have 9 options there. Second slot (the number being same as slot 1), you have only 1 option. Third slot can have any number other than the one in slot 1 and 2, giving you only 8 options. So you get 9*1*8 = 72 Note, you can have these numbers in 3 orders. Hence multiply 72 by 3. Did that help? Is it any easier for you now? Blueblu... thanks for the response. What I don't understand is that if we had all three distinct digits XYZ, we would have simply multiplied by 9*8*7. But XYZ can also be arranged in the following 6 ways but we don't multiple X*Y*Z with 6. XYZ XZY YXZ YZX ZXY ZYX Why is that so???? You are confusing distinct digits with the number of options. All of your digits X Y and Z are one of the digits from 1 to 9. Say you pick 678 and compare it with 876; in the first slot you put 9 which takes care of all 9 options including the digits 6,7 and 8 and so forth. This is why you don't multiply 9*8*7 with 3. I agree with GymArt  read Karishma's bible to get a clarity of your concepts. On it Thanks




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