Just took a test and it’s a pain in the ass finding individual problems via google. Want to check if I did this right:
Take normal deck of cards (shuffled), then start laying them out one by one. What is probability that all of the reds and blacks are next to each other (i.e 26 straight black, then 26 straight red, or vice versa).
Seems like a factorial situation, but I simply did:
2* [(26/52)(25/51)(24/50)…*(1/26)]
(with the inner brackets being for either reds or blacks, and the times 2 on account of being able to do it two different ways, reds then blacks, or blacks then reds). Seems proper.
One more: how many distinct (non-repeating) numbers are there between 100-999? I did:
(1098) / 3!
(with 10 possibilites in the first number, 9 in the second, 8 in the third), divided by the number of ways to arrange them. Didn’t have calculator for the test (could have brought one, but didn’t think I’d need it), and doing it just now getting home, that works out to 120, which sounds good. At least it’s not a fraction, though I believe that sometimes occurs and you can just round down.
There are 26! ways to order the black cards, and the same number of ways to order the red cards. Thus, there are 26!26! ways to draw first 26 red cards, then 26 black cards. Double this to get all the ways to order one color first, then the other = 226!*26!
The probability then is (2*26!*26!)/52! = 4.033 x 10^-15, or about one in every 250 trillion attempts. That’s the easiest way i know to solve this…
The first question seems correct to me, though I would have probably done it slightly differently.
On the second question, is it supposed to be “How many numbers between 100 and 999 have 3 distinct digits?”?
If so, then you have only 9 choices for the first number, but also 9 for the second, and then 8 possibilities for the third. Multiplying those you get 648. There is no need to divide, unless numbers like 478 and 784 are to be considered the same.
For the cards one: First, there are 52! ways to arrange the cards in any order.
Now count how many ways it can be done with red and black segregated.
There are 26! ways to arrange the red cards, and 26! ways to arrange the black cards.
Once the reds and blacks are arranged amongst themselves, there are two ways to arrange them with respect to each other: Black then red, or red then black.
So there are 2(26!)(26!) arrangements with red and black segregated, giving a probability of 2(26!)(26!) / 52!
I’m not sure I understand your second question. What do you mean by, “distinct (non-repeating) numbers…between 100-999”?
There are 52! permutations (ordered arrangements) of the entire deck. There are 26! permutations of all the black cards and the same number of permutations of all the red cards. Hence there are 26!26! orders in which you can get all the black cards and then all the red cards, and an equal number of orders in which you can get all the red cards and then all the black cards. Thus the answer is (226!*26!)/(52!). This is just another way of writing what you wrote, except that I believe the final term in your product should have been (1/27).
A flaw here; there are only 9 possibilities for the first digit, not 10 (it’s 100-999, not 000-999). This then leaves nine possible digits for the second number, and eight for the third, giving you 998 = 648 three-digit numbers with no digits repeated.
Dividing by 3! gives you 108, the total number of three-digit combinations, i.e. 648 and 486 are considered the same combination, even though there are different numbers. i suspect you want the 648 answer, not 108.
Good deal, should just lose a point or two for the 26 vs. 27 mistake. I actually wrote it the way most of you have said to do it, but that answer came from the way I THOUGHT it was supposed to be done, whereas the way I wrote it came from my understanding of what was going on, which I trust more than my memory.
For the second one, to make it more clear, take the numbers 100-999. How many numbers are there between the two that don’t have repeating numbers. So 100 counts as a number, but 101 doesn’t because you have two ones. 102 is good, as are the others up to 110, which doesn’t count because it has two ones. The next number that would count is 120, since 110-119 all have a repeating one.
Damnit, I just realised that I am off on my answer. I did 1098, but it should have been 998 because the first slot can’t include a 0. Hopefully I’m not penalized to much if the rest is right.
K, CJJ beat me to my own mistake. Looks like I’m not going to get much credit for that one, but assuming I didn’t make any other unexpected mistakes, I look to have 7 of 8 answers correct.
That’s what I put, but I know it’s wrong. As mentioned, you can’t have a 0 in the first slot, so I’m pretty sure it’s (998). The big question now is whether the dividing by 3! is necessary. I hope it is, as it would give me way more credit for the problem. Off the top of my head, you lose ten numbers each 100 for the first two repeating (110-119, 220-229, etc…). Then you lose 10 per hundred for the first and third repeating, so 101, 202. Then another 10 per hundred for the second and third repeating, so that’s about 30 per hundred. That’s 270 you’re losing right there. Then take away the first 98 (1-99). So that’s 368 from the 999 you’re losing. Around the answer that CJJ gave, though I’m probably missing on something.
Yeah, one per hundred for all three repeating. That’s 9 more to the 368= 377 you lose. Probably did some wrong addition or counted a number or two that I shouldn’t have, but the answer is definitely not around 120.
I just saw this so I’m jumping in a bit late. Here’s another way to solve the black / red card separation probability.
You flip the first card over and this card can be either black or red making p=1 for this first event. The second card must match the first card. There are 51 remaining cards of which only 25 match the first card, so for this second event p=25 / 51. By this reasoning, the 3rd card matching the first is 24 / 50 and so on until we get to the 26th card where p= 1 / 27.
So calculating:
25!
51•50•49•48•47… •27
we get 4.03292 x 10[sup]-15[/sup] which is what everyone else said. Just thought I’d take a different approach to get the same answer.
