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

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

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New post 11 Nov 2018, 02:58
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



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

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New post 11 Nov 2018, 03:48
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.

BAAC= 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,
AABC= 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
BCAA= 9 x 8 x 9 x 1=648

Therefore,
648+648+648= 1944 (Option D)
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Re: Suppose a “Secret Pair” number is a four-digit number in which two adj  [#permalink]

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New post 14 Feb 2019, 23:24
Diwakar003 wrote:
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!


How XSSY is possible as question suggest that 2551 is not secret pair.please help
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Re: Suppose a “Secret Pair” number is a four-digit number in which two adj  [#permalink]

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New post 15 Feb 2019, 01:03
Sunshine1810 wrote:
Diwakar003 wrote:
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!


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


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!
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Re: Suppose a “Secret Pair” number is a four-digit number in which two adj   [#permalink] 15 Feb 2019, 01:03

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