Suppose a Secret Pair number is a four-digit number in which two adj

Question

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

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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

Mike

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Official Answer

D

joepc · Answer

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

10*3*72 = 2160

Hope it is correct, Fingers crossed :roll:

mikemcgarry · Answer

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 , 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 :-)

joepc · Answer

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 1800

Approach:-

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 Pair

Since 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 ways

Hope it is correct :-D

mikemcgarry · Answer

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 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 :-)

Diwakar003 · Answer

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!

vaibhavbhardwaj1987 · Answer

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

quantumliner · Answer

The answer is C (1800).

thefibonacci · Answer

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.

Beqa · Answer

Hi, I solved this question in less than 1 min with the correct answer. Here is my approach:
Choosing a "secret pair" number can be separated into 4 steps: 
STEP 1. Select the positions of the pair
I can choose where the pair is in the number. The number can be AABC, ABBC, or ABCC. There are 3 ways to do so. You can also notice that once we choose how the number looks like, we need to select 3 digits for A, B, and C.

STEP 2. Select A - The first digit:
A cannot = 0. Hence, there are 9 ways.

STEP 3. Select the B digit: 
As B is different from A. There are 9 ways (including 0)

STEP 4. Select the C digit:
As C is different from A and B. There are 8 ways.

Number of combination is the product of all possible options of each step: 3x9x9x8=1944.

eswarchethu135 · Answer

There exist three types of number cases to be solved.

i) _ _ 00
ii) 11_ _
iii) _00_

In case i)

For _ _00, The last 2 digits can be from 00,11,22,...99 => 10 possibilities
The first digit can be filled in 9 ways out of 10 numbers (0 to 9).
The second digit can be filled in 8 ways out of 10 numbers, in which 0 and the number filled in first digit's place should be excluded. So total 9∗8=72 ways

Now, consider _ _ 11, _ _ 22,...etc
The first digit can be filled in 8 ways as 0 can't be filled in the first digit and 1 also cannot be included. 
the second digit can be filled in 8 ways, as 0 can be included now but the digits in first and last position of the number should be excluded. So a total of 8∗8=64 ways.

These 64 ways are common for _ _11 to _ _ 99. So 64∗9=576 ways

Total possible numbers for case i) are  576+72=648  numbers

In case ii)

The first two digits can be from 11 to 99, 9 ways.
The third digit can be filled in 9 ways and the fourth digit can be filled in 8 ways. So total ways are 9∗8=72 ways.

Total possible numbers are  72∗9=648  numbers.

In case iii)

For _00_, the first digit can be filled in 9 ways and the last digit can be filled in 8 ways. Total ways are 9∗8=72 ways
For _11_, _22_, .... _99_ , The first digit can be filled in 8 ways (excluding 0 and digit in second place) and the last digit can be filled in 8 ways (including 0 and excluding 2 other digits already in the number). So total ways are 8∗8=64 ways for each possibility and there are 9 such possibilities, so total ways are 64∗9=576 ways.

Total possible numbers are  576+72=648  ways.

Finally, the total possible numbers are  648+648+648=1944  numbers.

OPTION :  D

EncounterGMAT · Answer

A really good question! Here's what I've done.

We know that a total of 10 nos.(0-9) are possible but first number can't be 0.

B AA C= 9(as 0 can't be the first number) x 9(as out of 10 nos., one no. is the first digit) x 1 (repeat of 2nd digit) x 8 (as 2 nos out of total 10 have already been filled out in first two digits) = 648

Similarly, 
 AA BC= 9(as 0 can't be the first number) x 1(as 0 can't be the first digit) x 9(remaining 9 nos) x 8= 648, and
BC AA = 9 x 8 x 9 x 1=648

Therefore,
648+648+648= 1944 (Option D)

Sunshine1810 · Answer

How XSSY is possible as question suggest that 2551 is not secret pair.please help

Diwakar003 · Answer

Hey  Sunshine1810

I think you misread the question. The question says 2552 is not a secret pair. Hope this helps! Please feel free to let me know if you still have any doubts.

Cheers!

SudsMeister · Answer

3 cases are possible

Repetition in spot 1 and 2, repetition in spot 2 and 3, and repetition in spot 3 and 4.

Case 1: Spot 1 and 2 repeat
9 options for spot 1 and 2 (0 is not possible as 4 digit number) x 9 options for spot 3 (1 used by spot 1 and 2 but 0 added) x 8 options for spot 4 (2 numbers used already) = 9 x 9 x 8 = 648 ways.

Case 2: Spot 2 and 3 repeat
This will have 2 sub cases:

A. When 0 is the repeating number:
9 options for spot 1 x 1 option for spot 2 and 3 (0 alone allowed) x 8 options for spot 4 (2 numbers used already) = 9 x 1 x 8 = 72

B. When other than 0 is the repeating number:
8 options for spot 1 (0 not allowed and 1 number taken by spot 2 and 3) x 9 options for spot 2 and 3 (0 is not allowed) x 8 options for spot 4 (2 numbers used already) = 8 x 9 x 8 = 576

Total for case 2 = A + B = 72 + 576 = 648 ways.

Intuitively we can see that case 3 will be same as case 2.
Case 3 = 648 ways.

Total = Case 1 + Case 2 + Case 3 = 648 + 648 + 648 = 1944 ways.

Answer D.

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