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Bunuel
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How many integers between 1000 and 9999 have exactly one pair of equal
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12 Feb 2021, 09:00
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46% (03:02) correct 54% (02:39) wrong based on 70 sessions

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How many integers between 1000 and 9999 have exactly one pair of equal digit such as 4049 or 9902 but not 4449 or 4040?

A. 3888
B. 3887
C. 3886
D. 3885
E. 3884


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How many integers between 1000 and 9999 have exactly one pair of equal
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12 Feb 2021, 10:10
Bunuel
How many integers between 1000 and 9999 have exactly one pair of equal digit such as 4049 or 9902 but not 4449 or 4040?

A. 3888
B. 3887
C. 3886
D. 3885
E. 3884

Show more

We want the digits in the form of AABC, BAAC etc then we randomly scramble the letters.

In any case, there are 9 ways to pick the first digit, then 9 ways to pick the 2nd digit (since we can include 0) and 8 ways to pick the 3rd digit. Also note AABC and AACB are the same cases, out of the 9*9*8 cases they each have an identical pair by swapping B and C, so divide by 2 to get rid of the duplicates.

Then there are 9*9*8/2 = 81*4 cases. Each case has 4!/2 = 12 scrambles and we get 81*4*12 = 81*48 = 3888.

Ans: A

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How many integers between 1000 and 9999 have exactly one pair of equal
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19 Nov 2022, 06:36
Another quick way out when you have this kind of answer choices is to realize that the correct choice has to be divisible by 12.

if 4!/2!=12.
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How many integers between 1000 and 9999 have exactly one pair of equal
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26 Aug 2024, 06:31
TestPrepUnlimited

Bunuel
How many integers between 1000 and 9999 have exactly one pair of equal digit such as 4049 or 9902 but not 4449 or 4040?

A. 3888
B. 3887
C. 3886
D. 3885
E. 3884


We want the digits in the form of AABC, BAAC etc then we randomly scramble the letters.

In any case, there are 9 ways to pick the first digit, then 9 ways to pick the 2nd digit (since we can include 0) and 8 ways to pick the 3rd digit. Also note AABC and AACB are the same cases, out of the 9*9*8 cases they each have an identical pair by swapping B and C, so divide by 2 to get rid of the duplicates.

Then there are 9*9*8/2 = 81*4 cases. Each case has 4!/2 = 12 scrambles and we get 81*4*12 = 81*48 = 3888.

Ans: A
Show more

The whole thing makes sense, 9*9*8 and the 12 scrambles

but what does this mean "Also note AABC and AACB are the same cases, out of the 9*9*8 cases they each have an identical pair by swapping B and C, so divide by 2 to get rid of the duplicates."

Bunuel TestPrepUnlimited KarishmaB

if there is 4408 and 4480 shouldn't it be considered as two cases? the question asks how many integers have a pair of same digits, and these two are different integers technically.


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How many integers between 1000 and 9999 have exactly one pair of equal
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27 Aug 2024, 00:53
Fiji2000

TestPrepUnlimited

Bunuel
How many integers between 1000 and 9999 have exactly one pair of equal digit such as 4049 or 9902 but not 4449 or 4040?

A. 3888
B. 3887
C. 3886
D. 3885
E. 3884


We want the digits in the form of AABC, BAAC etc then we randomly scramble the letters.

In any case, there are 9 ways to pick the first digit, then 9 ways to pick the 2nd digit (since we can include 0) and 8 ways to pick the 3rd digit. Also note AABC and AACB are the same cases, out of the 9*9*8 cases they each have an identical pair by swapping B and C, so divide by 2 to get rid of the duplicates.

Then there are 9*9*8/2 = 81*4 cases. Each case has 4!/2 = 12 scrambles and we get 81*4*12 = 81*48 = 3888.

Ans: A
The whole thing makes sense, 9*9*8 and the 12 scrambles

but what does this mean "Also note AABC and AACB are the same cases, out of the 9*9*8 cases they each have an identical pair by swapping B and C, so divide by 2 to get rid of the duplicates."

Bunuel TestPrepUnlimited KarishmaB

if there is 4408 and 4480 shouldn't it be considered as two cases? the question asks how many integers have a pair of same digits, and these two are different integers technically.


­
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­
They are considered as 2 different cases when you multiply the whole thing by 4!/2!. Since you are going to do that re-arrangement later, while selecting the digits, you cannot select "say B for second spot and C for third spot". You just say that you need to select 2 of the remaining 9 digits and you can do that in 9C2 = 9*8/2 ways.

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