xc630 said:
Hello I have to make a tree diagram showing the various possible outcomes for this problem:
Four cards numbered 1 through 4 are shuffled and 3 different cards are chosen.
So one possible outcome is 1,2,3 another is 1,3,4, and so on but how would I pput this in a diagram?
Another thought:
You have probably heard of permutaions and combinations, the numbers of arrangements and choices, or combinations, that n things can be arranged or chosen. We designate the permutations of n things taken n at a time as nPn and the permutations of n things taken r at a time as nPr where P stands for permutations, n stands for the number of things involved, and r is less than n. To find the number of permutations of n dissimilar things taken n at a time, the formula is nPn = n! which is n factorial which means n(n-1)(n-2)(n-3).......3x2x1. To find the number of permutations of n dissimilar things taken r at a time, the formula is nPr = n(n-1)(n-2)(n-3)..........(n-r+1).
We designate the combinations of n things taken n at a time as nCn and the combinations of n things taken r at a time by nCr. To find the number of combinations of n dissimilar things taken r at a time, the formula is nCr = n!/[r!(n-r)!] which can be stated as n factorial divided by the product of r factorial times (n-r) factorial. Example: In how many ways can a committee of three people be selected from a group of 12 people? We have 12C3 = (12!)/[3!(9!) = 220. How many different ways can you combine A, B, C, and D in sets of three? Clearly, 4C3 = (4x3x2x1)/(3x2x1)(1) = 4. How many handshakes will take place between six people in a room when they each shakes hands with all the other people in the room one time? Here, 6C2 = (6x5x4x3x2x1)/(2x1)(4x3x2x1) =15.
Lets look at our expression for combinations, nCr = n!/[r!(n-r)!], assuming 0! = 1. What is the number of possible combinations of 4 things taken 2 at a time? 4C2 = 4!/[2!(2!)] = 6. What about 4 things taken 1 at a time? 4C1 = 4!/[1!(3!)] = 4. What about 4 things taken 3 at a time? 4C3 = 4!/[3!(1!)] = 4. What about 4 things taken none at a time? 4C0 = 4!/[0!(4!)] = 1. What about 4 things taken 4 at a time? 4C4 = 4!/[4!(0!)] = 1. Lets try 3 things taken 2 at a time? 3C2 = 3!/[2!(1!)] = 3. How about 3 things taken 1 at a time? 3C1 = 3!/[1!(2!)] = 3. How about 3 things taking none at a time? 3C0 = 3!/[3!(0!)] = 1. How about 3 things taken 3 at a time? 3C3 = 3!/[3!(0!)] = 1. Do you recognize anything familiar about the resulting combinations of 3 things taken 0, 1, 2, and 3 at a time and 4 things taken 0, 1, 2, 3, and 4 at a time? Yes, the numbers of combinations for each series of conditions match the numbers in the 3rd and 4th rows of Pascals triangle, 1, 3, 3, 1 and 1, 4, 6, 4, 1. How about that? We can use Pascal's triangle to find any combination of n things taken r at a time by simply referring to the nth row of the triangle. In general, the sequence of numbers in the nth row represent the possible combinations of "n" things taken 0, 1, 2, 3, ...n at a time, in that order across the row..
Row
0..................................................................................1
1..............................................................................1......1
2...........................................................................1.....2.....1
3........................................................................1....3......3.....1
4.....................................................................1....4.....6......4.....1
5..................................................................1....5...10....10.....5.....1
6...............................................................1....6...15....20....15....6.....1
7............................................................1....7...21...35...35....21.....7.....1
8.........................................................1....8...28...56...70....56...28....8.....1
9......................................................1....9...36..84..126..126..84....36.....9.....1
10.................................................1...10..45.120..210.252..210..120...45...10....1
