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Magoosh GMAT Instructor
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Suppose a “Secret Pair” number is a fourdigit number in which two adj [#permalink]
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01 Mar 2017, 11:56
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Suppose a “Secret Pair” number is a fourdigit number in which two adjacent digits are equal and the other two digits are not equal to either one of that pair or each other. For example, 2209 and 1600 are “Secret Pair” numbers, but 1333 or 2552 are not. How many “Secret Pair” numbers are there?
(A) 720 (B) 1440 (C) 1800 (D) 1944 (E) 2160This is a somewhat easier problem from a list of 15 challenging problems for the GMAT Quant. For the whole collection, as well as the OE for this question, see: Challenging GMAT Math Practice QuestionsMike
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Re: Suppose a “Secret Pair” number is a fourdigit number in which two adj [#permalink]
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01 Mar 2017, 14:58
Thanks Mike for this Question, been a Fan of your questions many times. If I am not wrong, this question has a flaw, The words like Number and digits makes confusing If it is a Number Type it can't have 00 in the beginning If it is a Digits Type it can have 00 in the beginning(like a Secret digit like ATM pin, Etc..) If it is digit type the answer is E 2160Approach : Given There are _ _ _ _ 4 digits or 4 Positions named 1,2,3,4 respectively And the total set of Consecutive digits are (00,11,22,33,44,55,66,77,88,99) there are 10 such pairs The consecutive positions are 12, 23, and 34  there are 3 consecutive positions Ans : Choosing one consecutive digit which is 10 ways And these digits can go into any positions namely 12, 23, and 34 which is 3 ways And the remaining digit places can be filled with 8*9 = 72 ways 10*3*72 = 2160 Hope it is correct, Fingers crossed



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Re: Suppose a “Secret Pair” number is a fourdigit number in which two adj [#permalink]
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01 Mar 2017, 15:13
joepc wrote: Thanks Mike for this Question, been a Fan of your questions many times. If I am not wrong, this question has a flaw, The words like Number and digits makes confusing If it is a Number Type it can't have 00 in the beginning If it is a Digits Type it can have 00 in the beginning(like a Secret digit like ATM pin, Etc..) If it is digit type the answer is E 2160Approach : Given There are _ _ _ _ 4 digits or 4 Positions named 1,2,3,4 respectively And the total set of Consecutive digits are (00,11,22,33,44,55,66,77,88,99) there are 10 such pairs The consecutive positions are 12, 23, and 34  there are 3 consecutive positions Ans : Choosing one consecutive digit which is 10 ways And these digits can go into any positions namely 12, 23, and 34 which is 3 ways And the remaining digit places can be filled with 8*9 = 72 ways 10*3*72 = 2160 Hope it is correct, Fingers crossed Dear joepc, I'm happy to respond. My friend, to the best of my understanding, there is no flaw in this question. The question makes very clear that we are looking for a particularly category of numbers, fourdigit numbers. The digits are mentioned to describe the restraint, but clearly we want to find numbers. You interpreted it as if we were just looking for fourdigit expressions, such as 0027. This would be a possibility if the question were looking for something such as "sets of four digits," but for an actual number, this doesn't work. Does all this make sense? Mike
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Re: Suppose a “Secret Pair” number is a fourdigit number in which two adj [#permalink]
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01 Mar 2017, 21:09
one of the answer choice, which by the way is correct answer, is typed wrong.. I was breaking my head and wondering what is going on



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Re: Suppose a “Secret Pair” number is a fourdigit number in which two adj [#permalink]
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02 Mar 2017, 05:19
Thanks Mike for the clarification This is very good question, But it took some time to solve ; and a very Good Mind GYM question. Answer is C 1800Approach:Pairs { 00,11,22,33,44,55,66,77,88,99} 10 Pairs Treat the 00 Pair separate i.e Case I and Other pairs Together separate i.e Case II Case I : 00 PairSince 4 digits each digit place is named as position 1,2,3,4 respectively A) 00 pair in position 2 and 3 First Position has 9 Options to fill Fourth Position has 8 Options = 72 Ways B) 00 in position 3 and 4 First Position has 9 Options to fill Second Position has 8 Options = 72 Ways C) 00 in Position 1 and 2 cannot be considered as they are not Numbers Total for Case I = 72+72 = 144 pairs Case II Other Pairs{11,22,33,44,55,66,77,88,99} Take Any one pair let say 11 A)Put 11 in First and second position Third position can be filled in 9 ways Fourth Position can be filled in 8 ways = 72 ways B) Put 11 in second and third position First position can be filled in 8 ways Fourth Position can be filled in 7 ways = 56 Ways C) Put 11 in third and fourth position First position can be filled in 8 ways Second Position can be filled in 7 ways = 56 Ways Total = 72+56+56 = 184 ways This for 11 and for remaining 22,33,44,55,66,77,88,99 Total for case II 184*9 = 1656 Ways Total ways = Case I + case II = 144+1656 = 1800 waysHope it is correct



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Re: Suppose a “Secret Pair” number is a fourdigit number in which two adj [#permalink]
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02 Mar 2017, 11:28
Dear joepc, I'm happy to respond. My friend, I'm sorry to say that there are flaws in your method of counting. Look at this part Quote: Total for Case I = 72+72 = 144 pairs
Case II Other Pairs{11,22,33,44,55,66,77,88,99} Take Any one pair let say 11
A )Put 11 in First and second position, Third position can be filled in 9 ways, Fourth Position can be filled in 8 ways = 72 ways
B) Put 11 in second and third position, First position can be filled in 8 ways, Fourth Position can be filled in 7 ways = 56 Ways
C) Put 11 in third and fourth position, First position can be filled in 8 ways, Second Position can be filled in 7 ways, = 56 Ways
Total = 72+56+56 = 184 ways This for 11 and for remaining 22,33,44,55,66,77,88,99 Total for case II 184*9 = 1656 Ways Here, part (A) is correct. For part (B), suppose we have X11X. It's true, we can put only 8 choices in the first slot (not 0 or 1)suppose we put 5 there. Now we have 511X. At this point, we have eight digits left for the fourth place (not 1 or 5, but any of the others). The mistake in part (C) is similar. Your method undercounted the possibilities. Does this make sense? Mike
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Re: Suppose a “Secret Pair” number is a fourdigit number in which two adj [#permalink]
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02 Mar 2017, 13:18
Thanks Mike,
It was an eye opener, and It was a wonderful question, Keep Posting.
I got the answer, But not telling this time, waiting for your revelation
Long Live My friend Mike



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Re: Suppose a “Secret Pair” number is a fourdigit number in which two adj [#permalink]
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03 Mar 2017, 09:41
Three types of Secret pairs can be formed. Let X, Y, and S be different integers.
1. SSXY  648 nos
2. XYSS  648 nos 3. XSSY  648 nos
Hence the total numbers  1944. Hopefully this is correct.
Cheers!



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Suppose a “Secret Pair” number is a fourdigit number in which two adj [#permalink]
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05 Mar 2017, 20:41
There are 10 pairs {00,11,22,33,44,55,66,77,88,99}
and 3 locations where these pairs can be fitted
Case 1 : when 00 is the pair
there are just two locations where his pair can go 23 and 34
no of possible options for other two numbers = 9*8 so total no of possible numbers= 1*2*8*9= 144
Case 2
When pair other than 00 is chosen. Now this pair has three locations where it can be fitted 12,23,34
But Here again there can be two cases
A ) when the position of such a pair is 23 and 34
the number of options for other two digits is 8*8 because 0 can not come at 1st position
so total in such case is 9*2*8*8
B) When position of such pair is 12
The option for other two digits is 9*8 because 0 is also a valid number in this case total = 9*1*9*8
Total = 1*2*9*8+ 9*1*9*8+9*2*8*8 = 1944
So I think 1944 is the answer



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Re: Suppose a “Secret Pair” number is a fourdigit number in which two adj [#permalink]
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05 Mar 2017, 21:28
The answer is C (1800).
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Suppose a “Secret Pair” number is a fourdigit number in which two adj [#permalink]
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08 Sep 2017, 14:34
Mike, I'm curious here. I got the right answer, but only because I was able to eliminate A, B, C, and E. This question revealed an issue in my logic and I'm hoping you can help me determine why one method works and the other does not. This was my (erroneous) approach. A is the first digit I chose, then B is the second, then C is the third. AABC: (9 options)(1 option)(9 options)(8 options) = 648 BAAC: (8 options)(10 options)(1 option)(7 options) = 720 BCAA: (8 options)(7 options)(10 options)(1 options) = 720 Wrong answer: 2088 Now I picked D because of intuition, but after looking over the explanation, I'm very interested because in math you can often take more than one approach, as we did, but mine of course was incorrect . I see that when picking numbers for the three cases, I picked the digits from A to C, like so: AABC, BAAC, and BCAA. However you picked them going from left to right: AABC, ABBC, and ABCC. For example, for a "Secrect Number" such as BAAC, I got 720 options (8*10*7), whereas you got 648 (9*9*8) for the same type of "Secret Number", just you picked the first digit first (e.g. ABBC). So, what gives? What did my approach forget to account for? I was not under the impression that when choosing combinations, one had to go from left to right, but now I am wondering if that was the error in my logic. Or, perhaps I should start with the most restrictive conditions first, then, move to least restrictive?



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Suppose a “Secret Pair” number is a fourdigit number in which two adj [#permalink]
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08 Sep 2017, 15:24
brooklyndude wrote: Mike,
This was my (erroneous) approach. A is the first digit I chose, then B is the second, then C is the third. AABC: (9 options)(1 option)(9 options)(8 options) = 648 BAAC: (8 options)(10 options)(1 option)(7 options) = 720 BCAA: (8 options)(7 options)(10 options)(1 options) = 720
Wrong answer: 2088
Your fatal flaw comes with BAAC and BCAA. BAAC: (8 options)(10 options)(1 option)(7 options) is incorrect BAAC should be: (9 options [can use anything except 0])(9 options [the remainder of the first #s & 0])(1 option)(8 options [any of the remaining #s]) =648 BCAA should be: (9 options [can use anything except 0])(9 options [the remainder of the first #s & 0])(8 options [any of the remaining #s])(1 option) =648 648 * 3  1944



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Re: Suppose a “Secret Pair” number is a fourdigit number in which two adj [#permalink]
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08 Sep 2017, 16:18
subsauce wrote: brooklyndude wrote: Mike,
This was my (erroneous) approach. A is the first digit I chose, then B is the second, then C is the third. AABC: (9 options)(1 option)(9 options)(8 options) = 648 BAAC: (8 options)(10 options)(1 option)(7 options) = 720 BCAA: (8 options)(7 options)(10 options)(1 options) = 720
Wrong answer: 2088
Your fatal flaw comes with BAAC and BCAA. BAAC: (8 options)(10 options)(1 option)(7 options) is incorrect BAAC should be: (9 options [can use anything except 0])(9 options [the remainder of the first #s & 0])(1 option)(8 options [any of the remaining #s]) =648 BCAA should be: (9 options [can use anything except 0])(9 options [the remainder of the first #s & 0])(8 options [any of the remaining #s])(1 option) =648 648 * 3  1944 Completely understand that. My question is  why? I want to understand the theory why one method works and the other does not.



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Re: Suppose a “Secret Pair” number is a fourdigit number in which two adj [#permalink]
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19 Sep 2017, 14:41
Good question.
Let's assume the pair of equal digits in in the first 2 cases, hence 0 is not usable.
We'll have 9*1*9*8 options, which equals 648 numbers.
Now, if the pair is something like ABBC > 9*9*1*8 (since now we can use 0 in the middle)
If the pair is in the end, which will give a number like ABCC: 9*9*8*1



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Suppose a “Secret Pair” number is a fourdigit number in which two adj [#permalink]
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27 Feb 2018, 08:07
quantumliner wrote: The answer is C (1800). Your calculations are pretty structured, however the "Sub case 2" sections are wrong. Units may be equal to 0, hence there should be 576 instead 504.



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Re: Suppose a “Secret Pair” number is a fourdigit number in which two adj [#permalink]
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28 Feb 2018, 06:34
mikemcgarry wrote: Suppose a “Secret Pair” number is a fourdigit number in which two adjacent digits are equal and the other two digits are not equal to either one of that pair or each other. For example, 2209 and 1600 are “Secret Pair” numbers, but 1333 or 2552 are not. How many “Secret Pair” numbers are there?
(A) 720 (B) 1440 (C) 1800 (D) 1944 (E) 2160This is a somewhat easier problem from a list of 15 challenging problems for the GMAT Quant. For the whole collection, as well as the OE for this question, see: Challenging GMAT Math Practice QuestionsMike Hi Mike good question, the 2552 part had me thrown off a bit :p there would be 3 cases > 1) aabc 2) baac 3) bcaa aabc > 9*1*9*8 = 648 baac > 9*9*8*1 = 648 bcaa > 9*9*8*1 = 648 => 1944 = option d.
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Re: Suppose a “Secret Pair” number is a fourdigit number in which two adj [#permalink]
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28 Feb 2018, 19:52
Hello,
Basically just restating what has been said above:
1. Find the number of permutations using formula 10! / (10  7)! = 720. (10C3, but with ordering) 2. Subtract out the permutations that start with zero, which is 1/10th of the 720, or 72. 720  72 = 648 3. Recognize that there are 3 ways to do this (double number can be first, middle, or last). 648 * 3 = 1,944




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