2 votes 2 votes A box contains 6 red, 8 green, 10 blue, 12 yellow and 15 white balls. What is the minimum no. of balls we have to choose randomly from the box to ensure that we get 9 balls of same color? How to solve such problems using pigeonhole principle? Combinatory discrete-mathematics engineering-mathematics combinatory + – gauravkc asked Jan 20, 2018 gauravkc 2.2k views answer comment Share Follow See all 6 Comments See all 6 6 Comments reply Show 3 previous comments gauravkc commented Jan 20, 2018 reply Follow Share Awesome .. Thanks :) 1 votes 1 votes Mk Utkarsh commented Jan 21, 2018 reply Follow Share This is how i solve maybe can help what can be the worst case possible for not able to pick up 9 balls of same colour 1) picked up 8 white balls 2) picked up 8 yellow balls 3) picked up 8 blue balls 4) picked up 8 Green balls 5) picked up 6 Red balls till now we picked up 38 (8+8+8+8+6) balls and still don't have 9 balls of same colour and as soon as we pick the next one we meet the condition given so 39. 2 votes 2 votes MiNiPanda commented Jan 21, 2018 reply Follow Share Ashwin Kulkarni & Mk Utkarsh Great explanation.. But Ashwin can you please explain the last sentence which you said while using the formula.."..already we don't have 2 red balls..." Do you mean that picking up min of 41 balls means on an average picking 8.2 balls of each color. But since only 6 red balls are there, so considering additional two red balls is false. So subtract 2. But we also have only 8 green balls and not more than that. We don't we subtract another 1? I understood the 1st method..you wrote it really well! :) 0 votes 0 votes Please log in or register to add a comment.
