If a four-digit number is chosen at random what's the probability that it contains 2 pairs?

[Zelda22]

I'm confused by this question. The sum of my answers should be 1, So I think is 0.027
but I think I'm doing something wrong.

10 digits choice, choose 4, so there is 10^4 outcomes= 10 000

-All different-

a b c d
a- 10 choices
b- 9
c- 8
d- 7

10 x 9 x 8 x 7 = 5040/10 000= 0.504

- 1 pair-

a a b c

a1- 10 choices
a2- 1 (repeated)
b- 9
c- 8

10 * 1 * 9 * 8 = 720

4C2= 6
720 * 6 = 4320
4320/10 000= 0.432

- 2 pairs -
a a b b
a1- 10 choices
a2- 1
b1- 9
b2- 1

10 * 1 * 9 * 1 = 90
4C2= 6 , (6/ 2=3 , I'm not sure if this is correct)
3 * 90 = 270
270/10 000 = 0.027

- 3 the same-
a a a b
a1- 10 choices
a2- 1
a3- 1
b- 9

10 * 1 * 1 * 9 = 90
4C3 =4
4 * 90 =360
360/10 000= 0.036

- 4 the same-
a a a a
a1- 10 choices
a2- 1
a3- 1
a4- 1

10 * 1 * 1 * 1 =10
4C4= 1
1 * 10 = 10
10/ 10 000= 0.001

 

[Dr.Peterson]

- 2 pairs -
a a b b
a1- 10 choices
a2- 1
b1- 9
b2- 1

10 * 1 * 9 * 1 = 90
4C2= 6 , (6/ 2=3 , I'm not sure if this is correct)
3 * 90 = 270
270/10 000 = 0.027

Since the number 6 is small, you can check for yourself. List all numbers consisting of two1's and two 2's, for example.

Then think about how this count interacts with your count of 90.

Or, simply tell us your thinking about whether 6 or 3 is correct, perhaps your reasons for each option. You may convince yourself by trying to convince us!

 

[pka]

I'm confused by this question. The sum of my answers should be 1, So I think is 0.027
but I think I'm doing something wrong.

I am very confused. The wording is poor to say the lest.
We are to choose a four digit number. Does the mean [imath]1000\le X\le 9999[/imath] ?
The statement says contains two pair. [imath]2277~\&~3003[/imath] both work, they contain two pair.
BUT do [imath]7777\text{ or }0011[/imath] also work?


[imath][/imath][imath][/imath][imath][/imath][imath][/imath]

 

[Zelda22]

Since the number 6 is small, you can check for yourself. List all numbers consisting of two1's and two 2's, for example.

Then think about how this count interacts with your count of 90.

Or, simply tell us your thinking about whether 6 or 3 is correct, perhaps your reasons for each option. You may convince yourself by trying to convince us!

4!/2! = 6 [1122, 1212, 1221, 2211, 2121, 2112]

6 * 90= 540
540/10000= 0.054
this why I’m confused,
The sum of my answers should be 1.

so I was wondering since I have 2 doubles, 6/2=3

then, it works
3 * 90= 270
279/10000= 0.027

please explain

 

[Zelda22]

I am very confused. The wording is poor to say the lest.
We are to choose a four digit number. Does the mean [imath]1000\le X\le 9999[/imath] ?
The statement says contains two pair. [imath]2277~\&~3003[/imath] both work, they contain two pair.
BUT do [imath]7777\text{ or }0011[/imath] also work?


[imath][/imath][imath][/imath][imath][/imath][imath][/imath]

 

[pka]

Please do tell me what in the world that post has to do with what I asked?

 

[Dr.Peterson]

4!/2! = 6 [1122, 1212, 1221, 2211, 2121, 2112]

6 * 90= 540
540/10000= 0.054
this why I’m confused,
The sum of my answers should be 1.

so I was wondering since I have 2 doubles, 6/2=3

then, it works
3 * 90= 270
279/10000= 0.027

please explain

First, you really should have initially shown us the entire problem with its original wording, so we could be sure of the meaning; I had the same questions as @pka but held them back. The problem, as we now know, is:


The examples are helpful in showing that the pair can be in any positions, not together.

But this still doesn't make it quite clear whether the first digit can be 0, since it doesn't say four-digit number, but four-digit poker hand, whatever that is. Was this defined previously?

As for your thinking, you have the right thoughts, but need to actually think them! You need to try harder to convince yourself one way or the other, so you can have confidence in the future.

Checking the sum of your answers is good -- I am often not confident until such a check works. It gives you good reason to question the 6. So now you have to explain why you have to divide by 2! Can you expand on "since I have 2 doubles"?

Your 90 counts ways to choose two digits to use, in order. There's a first and a second. But your list of 6 ways to arrange pairs doesn't distinguish a first and a second, does it? How can you correct for that?

I might instead have calculated the number of ways to choose (non-ordered) pair of digits (combinations, not permutations), and then ways to arrange those two digits, perhaps by choosing the 2 places to put the larger of the two digits (and put the smaller digit in the remaining places).

If you get the same answer from both of these approaches, then you can be confident (at least if it still makes the total 1!).

 

[Steven G]

Is the number of ways to get one pair = 864? (assuming numbers can start with 0's)

 

[Zelda22]

I am very confused. The wording is poor to say the lest.
We are to choose a four digit number. Does the mean [imath]1000\le X\le 9999[/imath] ?
The statement says contains two pair. [imath]2277~\&~3003[/imath] both work, they contain two pair.
BUT do [imath]7777\text{ or }0011[/imath] also work?


[imath][/imath][imath][/imath][imath][/imath][imath][/imath]

Is the number of ways to get one pair = 864? (assuming numbers can start with 0's)

For 1 pair this is what I did.
a a b c

a1- 10 choices
a2- 1 (repeated)
b- 9
c- 8

10 * 1 * 9 * 8 = 720

4C2= 6
720 * 6 = 4320
4320/10 000= 0.432

 

[Steven G]

Is the number of ways to get one pair = 864? (assuming numbers can start with 0's)

I meant 8640.