Help: Perms and combs (1 Viewer)

leehuan · Oct 4, 2015

Can't find the trick.

$From a pack of nine cards numbered 1, 2, 3, ..., 9, three cards are drawn at random and laid on a table from left to right. \\ What is the probability that the digits are drawn in descending order?$

Carrotsticks · Oct 4, 2015

Hint: For every choice of three cards, there exists exactly one permutation where the numbers are descending.

Drsoccerball · Oct 4, 2015

leehuan said:
Can't find the trick.

$From a pack of nine cards numbered 1, 2, 3, ..., 9, three cards are drawn at random and laid on a table from left to right. \\ What is the probability that the digits are drawn in descending order?$

Is the answer 1/4

InteGrand · Oct 4, 2015

Drsoccerball said:
Is the answer 1/4

Simply 1/3! = 1/6.

leehuan · Oct 4, 2015

Carrotsticks said:
Hint: For every choice of three cards, there exists exactly one permutation where the numbers are descending.

Wow. Golden. Thanks.

$\frac { ^{ 9 }{ { C }_{ 3 } } }{ ^{ 9 }{ { P }_{ 3 } } }$

InteGrand · Oct 4, 2015

leehuan said:
Wow. Golden. Thanks.

$\frac { ^{ 9 }{ { C }_{ 3 } } }{ ^{ 9 }{ { P }_{ 3 } } }$

Or, given any hand, there's 3! = 6 ways to order them, and exactly one is in descending order, so 1/6.

leehuan · Oct 4, 2015

InteGrand said:
Or, given any hand, there's 3! = 6 ways to order them, and exactly one is in descending order, so 1/6.

Yep that makes sense too. I completely overlooked that only one option out of 6 is in descending order.

leehuan · Oct 6, 2015

Just going to post my next question in this original thread..
________________________________________________
My friend gave me this question and I'm stumped again.
$How many combinations of 4 digits numbers exist where the order of the digits is not important. (i.e. 4040 is the same as 4004)$

Oh and 0 can be the first number.

rand_althor · Oct 6, 2015

Do you know the answer? I'm getting 5860.

leehuan · Oct 6, 2015

Nope, he made it up.

Sorry also 0 can be the first number. But otherwise mind showing working?

kawaiipotato · Oct 6, 2015

I got 549 but I'm not confident with this. nevermind, it's definitely wrong if 0 can be the first number
I got 670 including 0 as first number

rand_althor · Oct 6, 2015

Oh, well here's what I did:
Case 1: 4 unique numbers (1234)
10*9*8*7 = 5040. 10 numbers for the first (0-9), 9 for the second (0-9 excluding the first number) etc.
Case 2: 2 unique numbers (1123)
10*1*9*8 = 720
Case 3: 1 unique numbers (1112)
10*1*1*9 = 90
Case 4: 0 unique numbers (1111)
10*1*1*1 = 10

In total: 5040 + 720 + 90 + 10 = 5860 combinations.

Edit: Didn't know 0 could be the first number. Fixed now.

leehuan · Oct 6, 2015

rand_althor said:
Oh, well here's what I did:
Case 1: 4 unique numbers (1234)
9*9*8*7 = 4536. 9 numbers for the first (1-9, assuming 0123 is not a 4 digit number), 9 for the second (0-9 excluding the first number) etc.
Case 2: 2 unique numbers (1123)
9*1*8*7 = 504
Case 3: 1 unique numbers (1112)
9*1*1*8 = 72
Case 4: 0 unique numbers (1111)
9*1*1*1 = 9

In total: 4536 + 504 + 72 + 9 = 5121 combinations.

Hm, so if 0 can be the first number would you just replace 9->10, 8->9 and 7->8?

And 2 x 2 unique numbers what about those?

rand_althor · Oct 6, 2015

leehuan said:
Hm, so if 0 can be the first number would you just replace 9->10, 8->9 and 7->8?

And 2 x 2 unique numbers what about those?

Yep I changed it.

My bad, didnt consider that case:

Case 5: 2 pairs of unique numbers (1122)
10*1*9*1 = 90

Total ways = 5860 + 90 = 5950 combinations.

leehuan · Oct 6, 2015

Coolio. I'll assume that is true until if someone like InteGrand corrects you. Thanks

leehuan · Oct 6, 2015

Hang on. If the order doesn't matter, take case 1 is there any need to divide by 4! ?

Zlatman · Oct 6, 2015

I got 715, I think the 5950 accounts for the order of the number, which we don't need to worry about.

No repeated numbers: 10C4 = 210 (pick any 4 from the 10)
2 unique numbers: 9C2 x 10 = 360 (pick one from the 10 to have 2 digits, then pick 2 from the remaining 9)
1 unique number: 10 x 9 = 90 (pick one from the 10 to have 3 digits, then pick one of the remaining 9)
No unique number: 10 = 10 (pick one of the 10 to have 4 digits)
Two pairs of repeated numbers: 10C2 = 45 (pick two of the 10 to have 2 digits)

Total numbers = 715

InteGrand · Oct 6, 2015

leehuan said:
Just going to post my next question in this original thread..
________________________________________________
My friend gave me this question and I'm stumped again.
$How many combinations of 4 digits numbers exist where the order of the digits is not important. (i.e. 4040 is the same as 4004)$

Oh and 0 can be the first number.

$If I have interpreted this correctly, you are simply asking for how many ways we can choose numbers from $0,1,...,9$ if we are to choose 4 in total (order unimportant). By ways, I mean we choose the number of ways $n_j$ of digit $j$ ($j = 0,1,...,9$) subject to the following conditions: $n_j \geq 0$ for all $j$ and $\sum _{j = 0}^9 n_j = 4$.If this is the case, then the answer is $\frac{13!}{4!\cdot 9!}=715$, using the stars and bars method.$

$(So \textsl{ways} are like: two 0's and two 1's, or one 2 and three 4's, etc. Order is unimportant, only important thing is how many of each digit is chosen.)$

leehuan · Oct 6, 2015

InteGrand said:
$If I have interpreted this correctly, you are simply asking for how many ways we can choose numbers from $0,1,...,9$ if we are to choose 4 in total (order unimportant). By ways, I mean we choose the number of ways $n_j$ of digit $j$ ($j = 0,1,...,9$) subject to the following conditions: $n_j \geq 0$ for all $j$ and $\sum _{j = 0}^9 n_j = 4$.If this is the case, then the answer is $\frac{13!}{4!\cdot 9!}=715$, using the stars and bars method.$

$(So \textsl{ways} are like: two 0's and two 1's, or one 2 and three 4's, etc. Order is unimportant, only important thing is how many of each digit is chosen.)$

Would not have thought about using stars and bars for this. That interpretation is correct. But how come it doesn't agree with Ziatman?

Zlatman · Oct 6, 2015

leehuan said:
Would not have thought about using stars and bars for this. That interpretation is correct. But how come it doesn't agree with Ziatman?

My answer was wrong, but I fixed it. For the last case, I assumed there was order (which I was correcting rand for, lol).

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