There are 16 marbles numbered 1 to 16 in a box. Marbles 1 to 5 are red, marbles
ID: 1941311 • Letter: T
Question
There are 16 marbles numbered 1 to 16 in a box. Marbles 1 to 5 are red, marbles 6 to 8 are green, and marbles 9 to 16 are blue. I draw out four marbles, one at a time without replacing them, and record the result as an ordered list of four color/number combinations. For example, I write R4, G6, R1, B10 if the marbles drawn are, in order, #4, #6, #1, #10.(a) How many possible results are there?
(b) Of these, for how many are both the first and last marbles red?
(c) For how many are the first and second marbles different colors?
(d) For how many are all four marbles the same color?
Explanation / Answer
You note how many available marbles there are for each draw, and multiply them together. 1) Total possibilities = 16 * 15 * 14 * 13 all 16 available for first, then one less for each subsequent draw. 2) First and last red: [CORRECTED] This is the same as number of cases where first and second are red, which is 5 * 4 * 14 * 13. It works out the same as the sum of (x = Blue or Green) R R R R: 5 * 4 * 3 * 2 R R x R: 5 * 4 * 11 * 3 R x R R: 5 * 11 * 4 * 3 R x x R: 5 * 11 * 10 * 4 In each case there is 5 * 4 which is multiplied by (3 * 2 + 3 * 11 + 11 * 3 + 11 * 10) = 3 * (2 + 11) + 11 * (3 + 10) = 3 * 13 + 11 * 13 = 14 * 13. 3) Different colors on first two draws: need to figure each color combination separately R B: 5 * 8 B R : 8 * 5 R G: 5 * 3 G R: 3 * 5 B G: 8 * 3 G B: 3 * 8 add those up and multiply by 14 * 13 which is the number of ways of drawing 2 more after you have drawn the first two: (80 + 30 + 48 = 158) * 14 * 13 Alternative method: Count R R = 5 * 4 B B = 8 * 7 G G = 3 * 2 20 + 56 + 6 = 82 Add those up, multiply by 14 * 13 and subtract from the answer to part 1) (16 * 15 - 82 = 158) * 14 * 13 (same answer) 4) All same color: Can't be green, since only 3 of those Red: 5 * 4 * 3 * 2 Blue: 8 * 7 * 6 * 5 multiply out and add
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