40 Slips are placed in a hat....

[veronica1999]

Forty slips are placed into a hat, each bearing a number 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10 with each number entered on four slips. Four slips are drawn from the hat at random and without replacement. Let http://www.artofproblemsolving.com/Forum/code.php?hash=516b9783fca517eecbd1d064da2d165310b19759&sid=c2203dbdfd142396d8e0d12f81599d60 be the probability that all four slips bear the same number. Let http://www.artofproblemsolving.com/Forum/code.php?hash=22ea1c649c82946aa6e479e1ffd321e4a318b1b0&sid=c2203dbdfd142396d8e0d12f81599d60 be the probability that two of the slips bear a number http://www.artofproblemsolving.com/Forum/code.php?hash=86f7e437faa5a7fce15d1ddcb9eaeaea377667b8&sid=c2203dbdfd142396d8e0d12f81599d60 and the other two bear a number http://www.artofproblemsolving.com/Forum/code.php?hash=d6bdb677e4af59f78f088681f60ac28a0d6b031b&sid=c2203dbdfd142396d8e0d12f81599d60. What is the value of http://www.artofproblemsolving.com/Forum/code.php?hash=bf91cc5da690d4285591cb473adfc911d2794d13&sid=c2203dbdfd142396d8e0d12f81599d60?

For p there are only 10 cases when all the numbers are the same.

For q there would be a lot more cases.
First, there will be 10C2 ways to choose the two numbers then i have to think of the ways the number can be arranged. xxyy can be arranged in 6 ways and then i have to multiply by two because the number x and y can be switched.

So my answer is 10C2X6X2/40C4 over 10/40C4

I am not sure why my answer is wrong.(Worried)

 

[I like Serena]

Forty slips are placed into a hat, each bearing a number 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10 with each number entered on four slips. Four slips are drawn from the hat at random and without replacement. Let http://www.artofproblemsolving.com/Forum/code.php?hash=516b9783fca517eecbd1d064da2d165310b19759&sid=c2203dbdfd142396d8e0d12f81599d60 be the probability that all four slips bear the same number. Let http://www.artofproblemsolving.com/Forum/code.php?hash=22ea1c649c82946aa6e479e1ffd321e4a318b1b0&sid=c2203dbdfd142396d8e0d12f81599d60 be the probability that two of the slips bear a number http://www.artofproblemsolving.com/Forum/code.php?hash=86f7e437faa5a7fce15d1ddcb9eaeaea377667b8&sid=c2203dbdfd142396d8e0d12f81599d60 and the other two bear a number http://www.artofproblemsolving.com/Forum/code.php?hash=d6bdb677e4af59f78f088681f60ac28a0d6b031b&sid=c2203dbdfd142396d8e0d12f81599d60. What is the value of http://www.artofproblemsolving.com/Forum/code.php?hash=bf91cc5da690d4285591cb473adfc911d2794d13&sid=c2203dbdfd142396d8e0d12f81599d60?

For p there are only 10 cases when all the numbers are the same.

For q there would be a lot more cases.
First, there will be 10C2 ways to choose the two numbers then i have to think of the ways the number can be arranged. xxyy can be arranged in 6 ways and then i have to multiply by two because the number x and y can be switched.

So my answer is 10C2X6X2/40C4 over 10/40C4

I am not sure why my answer is wrong.

Hi veronica1999! :)

You multiplied by 2 because x and y could be switched.
However, you should divide by 2 instead.

 

[veronica1999]

Hi veronica1999! :)

You multiplied by 2 because x and y could be switched.
However, you should divide by 2 instead.

Thanks for your help but the answer is 162. i think the answer multiplied another 6.
~Veronica

 

[I like Serena]

Thanks for your help but the answer is 162. i think the answer multiplied another 6.
~Veronica

Ah yes, you also forgot to take into account that you need a 2nd slip with the same number as the first for 3 choices, and also a 2nd slip with the same number as the second.I can see that this would be confusing, so let's take this apart a bit further.
You have ignored ordering from the start, and you have tried to compensate for that.
Let's keep the ordering.
That is, let's assuming each slip can be distinguished.
So we have 1a, 1b, 1c, 1d as the slips with the number 1.

It means you have a total of 40 x 39 x 38 x 37 choices.

If the 4 slips are the same, you have 40 choices for the first slip, 3 for the second, 2 for the third, and 1 for the last.
This is 40 x 3 x 2 x 1 choices.

$p = {40 \cdot 3 \cdot 2 \cdot 1 \over 40 \cdot 39 \cdot 38 \cdot 37} = {10 \over 40C4}$

As you can see, this is the same as what you already had.

With the pattern xxyy, you have 40 choices for the first, 3 choices for the second, 36 choices for the third, and 3 choices for the last slip.
This is 40 x 3 x 36 x 3 choices.

It can be shuffled in 4! = 24 ways, but since you can exchange x and x, y and y, and xx and yy, you need to divide by 2 three times.
More specifically, if you enumerate them, you'll be counting each of these combinations twice.
If you doubt this, you could perhaps enumerate them for say 3 numbers instead of 10.
Anyway, effectively you can only shuffle in $4! / 2^3 = 3$ ways.

$q = {40 \cdot 3 \cdot 36 \cdot 3 \times 3 \over 40 \cdot 39 \cdot 38 \cdot 37}$

$q / p = 162$

 

[veronica1999]

Ah yes, you also forgot to take into account that you need a 2nd slip with the same number as the first for 3 choices, and also a 2nd slip with the same number as the second.I can see that this would be confusing, so let's take this apart a bit further.
You have ignored ordering from the start, and you have tried to compensate for that.
Let's keep the ordering.
That is, let's assuming each slip can be distinguished.
So we have 1a, 1b, 1c, 1d as the slips with the number 1.

It means you have a total of 40 x 39 x 38 x 27 choices.

If the 4 slips are the same, you have 40 choices for the first slip, 3 for the second, 2 for the third, and 1 for the last.
This is 40 x 3 x 2 x 1 choices.

$p = {40 \cdot 3 \cdot 2 \cdot 1 \over 40 \cdot 39 \cdot 38 \cdot 27} = {10 \over 40C4}$

As you can see, this is the same as what you already had.

With the pattern xxyy, you have 40 choices for the first, 3 choices for the second, 36 choices for the third, and 3 choices for the last slip.
This is 40 x 3 x 36 x 3 choices.

It can be shuffled in 4! = 24 ways, but since you can exchange x and x, y and y, and xx and yy, you need to divide by 2 three times.
More specifically, if you enumerate them, you'll be counting each of these combinations twice.
If you doubt this, you could perhaps enumerate them for say 3 numbers instead of 10.
Anyway, effectively you can only shuffle in $4! / 2^3 = 3$ ways.

$q = {40 \cdot 3 \cdot 36 \cdot 3 \times 3 \over 40 \cdot 39 \cdot 38 \cdot 27}$

$q / p = 162$

Thank you!(Smile)