Question

1) Consider 3 bags of colored balls. Bag A contains a blue, a yellow and a green ball. Bag B contains a cyan and a magenta ball. Lastly, bag C contains a red, a black, a white and a grey ball.

If you decide to grab 1 ball from bag A, 1 ball from bag B and 2 balls from bag C, how many ways can you obtain 4 colored balls? Note that none of the colors mentioned are the same so there is no worry of obtaining duplicate colors.

2) Suppose you have 4 bars and 20 stars.

(a) Find the number of ways you can arrange them in a straight line.

(b) Explain, in your own words, how you can use part (a) to solve the question: “Find the number of solutions to the equation x1 + x2 + x3 + x4 + x5 = 20 where each xi are non-negative integers”

Answer 1

1. Total number of ways to obtain 4 colored balls is

= now of ways of selecting 1 from A * no of ways of selecting 1 from B * no of ways of selecting 2 from c

= 3 * 2 * 6

= 36

2. a) The total number of ways in which 20 stars and 4 bars can be arranged in a straight line is

b) Lets suppose the sum of 20 represents 2o stars. We need to divide these 20 stars into 5 parts(each part non-negative). After separating these 20 stars into 5 parts, we put bars in between those 5 parts. Since there are 5 parts, hence only 4 bars will be needed to separate them. The bars will be placed in between the stars. The 20 stars have 19 spaces between them. So, the total number of ways in which 4 bars can be distributed in those 19 spaces between the 20 stars will give the total number of non-negative solutions to the equation

x1 + x2 + x3 + x4 + x5 = 20

Required number of non-negative solutions is

= Number of ways of distributing 4 bars in 19 spaces

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