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

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

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

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

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

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!

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

The answer is C (1800).

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?

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.

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

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.

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.

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

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.

Hey, the "why" is because in your earlier method you were overlapping stuff!

For the _ X X _ case for instance... if X is 0 then you can put 9 digits on first slot and 8 on second. If X is NOT 0 however, then you can put 8 digits on first slot and 8 on last.

MOREOVER, I don't understand why 8*10*7 = 720?? You clearly made a typo somewhere. I was super confused because your answer as you wrote it should underestimate, not overestimate the options.

So, the best way to approach this exercise using your approach would be to realize that for the first case we have 9*1*9*8, for the second 9*9*1*8 and for the last 9*8*9*1 hence 648*3 = 1944.

0 _ _ _ = 2 x 9 x 8 ways = 144 (the position for adjacent ones can be selected in 2 ways)

0 0 _ _ = 9 x 8 ways = 72

Total combinations = 3 x 10 x 9 x 8 = 2160 (the position for adjacent ones can be selected in 3 ways)

Acceptable combination = 2160 - 144 - 72 = 1944