Batch 3 - Class 129 - Pascal Triangle Continued

• In how many ways can you distribute 10 identical balls into 4 distinct bags colored red, yellow, orange and green. You may place any number of balls in each bag, or leave a bag empty
• Answer: Consider the words formed by 10 B's and 3 P's (P being a partition where one bag ends and the next begins) - 13C3

• Revise correspondence with nCm
• Any number if sum of numbers above it on left and right
• Total of each row 2^n
• Lateral Symmetry
• Edges go from 1,1,1,1 to 1,2,3,4,5 to 1,3,6,10 (Triangular numbers) and so on

• Lets draw this slightly differently and see more patterns

• Hockey Stick Pattern - Example 1+3+6+10+15 = 35 - Why?
• Parallelogram Pattern - Example - the whole block from left top to 21, totals to 84 - Why? Successive Hockey sticks! or can do it recursively.
• Lets try one more way of drawing it

• What are the diagonal totals? Does anyone recognize these numbers? Can you think why?
• Can you spot them on the original Pascal triangle?
• There are several other advanced patterns hidden in Pascal triangle - such as value of e, pi, star of david, catalan numbers...
• (Shakuntala - 23) How many squares does a 3x3 grid have (Answer: 14). How many do 1x1, 2x2, 3x3, 4x4, 5x5... grids have (Answer: 1, 5, 14, 30, 55, ...). What is the pattern in these numbers?
• Answer: Sum of squares of first n numbers). Why do we land up with such a pattern (Answer, counting from largest to smallest square, there are 1, 2, 3, 4, 5... choices on how to select an edge of that size of square square on each side of the grid
• Instructor Note: Kids tend to be able to answer the sequence. For pattern, walk through number of largest squares to smallest in a given grid, and then prompt students to identify the pattern. Contrast with prior problem of number of rectangles in a 3x3 grid.
• (MC Chap 11 - Prob 9) 10 points are marked on one straight line, and 11 points on another parallel straight line. How many triangles and quadrilaterals can one construct using these 21 points. (Answer: 10. 11C2 + 11. 10C2 [Could also do as 21C3 - 11C3 - 10C3]; 10C2 . 11C2)
• (MC Chap 11 - Prob 56) How many ways are there to choose 6 cards from a complete deck, so that all 4 suites are present (Answer: 4.13C3.13^3 + 4C2. 13C2^2. 13^2)

• (CombProbs Problem 52) Can you arrange numbers 1..9 in a circle so that sum of two neighbors is never divisible by 3, 5 or 7
• Answer: 5,6,2,9,4,7,1,3,8,(5) - One way to do this is to figure out all combinations which are "allowed", i.e. not divisible by 3,5,7. Then start to work with constraints like number 2 can only go with 6 and 9, and build from there. Number 4 can only go with 9 and 7, so that extends the sequence to 6,2,9,4,7, and so on.