a pretty simple formula for figuring out the binomial coefficients (Dr. Math, 4): n! [n:k] = -------- k! (n-k)! 6 * 5 * 4 * 3 * 2 * 1 For example, [6:3] = ------------------------ = 20. 3 * 2 * 1 * 3 * 2 * 1 The triangular numbers and the Fibonacci numbers can be found in Pascal's triangle. The triangular numbers are easier to find: starting with the third one on the left side go down to your right and you get 1, 3, 6, 10, etc (Swarthmore, 5) 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 The Fibonacci numbers are harder to locate. To find them you need to go up at an angle: you're looking for 1, 1, 1+1, 1+2, 1+3+1, 1+4+3, 1+5+6+1 (Dr. Math, 4). Another thing I found out is that if you multiply 11 x 11 you will get 121 which is the 2nd line in Pascal's Triangle. If you multiply 121 x 11 you get 1331 which is the 3rd line in the triangle (Dr. Math, 4). If you then multiply 1331 x 11 you get 14641 which is the 4th line in Pascal's Triangle, but if you then multiply 14641 x 11 you do not get the 5th line numbers. You get 161051. But after the 5th line it doesn't work anymore (Dr. Math, 4). Another example of probability: Say there are four children Annie, Bob, Carlos, and Danny (A, B, C, D). The teacher wants to choose two of them to hand out books; in how many ways can she choose a pair (ladja, 4)? 1.A & B 2.A & C 3.A & D 4.B & C 5.B & D 6.C & D There are six ways to make a choice of a pair. If the teacher wants to send three students: 1.A, B, C 2.A, B, D 3.A, C, D 4.B, C, D If the teacher wants to send a group of "K" children where "K" may range from 0-4; in how many ways will she choose the children K=0 1 way (There is only one way to send no children) K=1 4 ways ( A; B; C; D) K=2 6 ways (like above with Annie, Bob, Carlos, Danny) K=3 4 ways (above with triplets) K=4 1 way (there is only one way to send a group of four) The above numbers (1 4 6 4 1) are the fourth row of numbers in Pascal Triangle (Ladj...
