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Suppose a “Secret Pair” number is a four-digit 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) 2160

This 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 Questions

Re: Suppose a “Secret Pair” number is a four-digit number in which two adj [#permalink]

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01 Mar 2017, 13:58

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 2160

Approach :-

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

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 2160

Approach :-

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

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, four-digit 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 four-digit 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
_________________

Mike McGarry Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

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
_________________

Mike McGarry Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

Suppose a “Secret Pair” number is a four-digit number in which two adj [#permalink]

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08 Sep 2017, 13:34

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?

Suppose a “Secret Pair” number is a four-digit number in which two adj [#permalink]

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08 Sep 2017, 14:24

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

Re: Suppose a “Secret Pair” number is a four-digit number in which two adj [#permalink]

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08 Sep 2017, 15:18

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.