There are 9 cards. Separate them into 3 stacks, each stack has 3 cards. Then, take away the cards, one at a time. You can only take card from the top. So, how many ways are there to take away all the 9 cards?
Please explain how you solve it. Thank you.
2007-07-18 16:09:23 · 6 answers · asked by nomade_contabile 1 in Science & Mathematics ➔ Mathematics
It helps me if I actually do it. But you can do it in your head too. Make the card piles A, B, and C. Now put those letters into patterns with nine letters. For example:
AAABBBCCC
ABCABCABC
CCCBBBAAA
CBACBACBA
Hope it helps.
2007-07-18 16:14:40 · answer #1 · answered by Leanna 3 · 0⤊ 0⤋
you have 9 cards
seperate into 3 stacks
each stack has 3 cards
*1st stack = 3 cards
*2nd stack = 3 cards
*3rd stack = 3 cards
You have only 1 way to take all the 9 cards because as you said you can only take the card from the top.
But if you will ask how many combination of cards you will get from the 3 stacks of cards with 3 cards in each stacks, the answer is 27. That is 3x3x3=27.
2007-07-18 16:43:46 · answer #2 · answered by beth 2 · 0⤊ 1⤋
The cards in each stack have to be drawn in order top to bottom. It's just a matter of how you switch back and forth between stacks until all the cards are drawn. Consequently, because the order of each stack is fixed, the problem reduces to one of combinatorics.
Number of ways = 9! / [3! * 3! * 3!] = 1680
2007-07-18 17:44:43 · answer #3 · answered by Northstar 7 · 0⤊ 0⤋
Leanna's answer is the best... the other two are wrong.
To figure out how ways there are to combine 9 letters with 3 A's, 3 B's, and 3 C's, (in which case, the order of the A's doesn't matter, the order of the B's doesn't matter, and the order of the C's doesn't matter), you do what McGraw did, but then divide by 3!=3*2*1=6 to avoid counting the different permutations of A's as different combinations, divide by 3! again to take care of the B's and again for the C's.
So the answer is 9!/(3!3!3!)=1680
2007-07-18 16:31:17 · answer #4 · answered by olhenry56 2 · 0⤊ 0⤋
There are three cards you can choose from each time, until you get to the end. Eventually you'll have two stacks, and then one. So there are three choices the first time, three the second time, and so on. Here's the number of choices for each draw, until there's only one left:
3 x 3 x 3 x 3 x 3 x 3 x 3 x 2 x 1 = 4374
2007-07-18 16:16:42 · answer #5 · answered by Billy 2 · 0⤊ 1⤋
The first card you remove could be any one of nine. The second has to be one of the remaining eight. The third, one of seven....
9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 9! = 362,880.
9! is called "nine factorial". This is actually a "permutation" question, not a probability question.
2007-07-18 16:19:11 · answer #6 · answered by Boots McGraw 5 · 1⤊ 1⤋