Posted 14 Jun 2014 Fifty school children lined up at random in five rows and ten columns. No two of the children have the same birthday. What is the probability that the youngest child among the oldest children in each column is NOT older than the oldest child among the youngest children in each row ? 0 Share this post Link to post Share on other sites

0 Posted 16 Jun 2014 The youngest child among the oldest children in each column is NEVER younger than the oldest child among the youngest children in each row. He is either older or of the same age. The same age happens when the youngest in any row is the oldest in that column. Number of ways in which the ages could be same can be calculated as below: Choose any place (chosen cell) in the 5x10 matrix. This can be done in 50 ways The kid in this cell must be the youngest in the row and oldest in the column. All other places can be filled randomly. So, there are 9 (in row) and 4(in column) "special cells" where combinations must be made. For other cells, the rest can be placed in 36! Ways. The 9 chosen kids in rows and 4 chosen in column can be palced in 9!*4! Ways. Now the only thing that remains to do is to choose 1 kid for the chosen cell and 9 kids for the rows and 4 kids for the columns in the special cells. For any given kid lets say there are x kids younger than him/her and 49-x elder kids Now we need to choose 4 kids out of x younger kids and 9 kids out of 49-x elder kids. Note that x varies from 4 to 40 ONLY. ^{4}C_{4} * ^{45}C_{9} + ^{5}C_{4} * ^{44}C_{9} + .... + ^{40}C_{4} * ^{9}C_{9} Using excel for this, the total combinations comes out to 9.378E + 11 Now, the total possible ways of placing the kids is 50! Total ways in which the age is same is = 50 * 36! * 4! * 9! * the above combinations Relying on excel getting the calculations right, there is 1 in 200 times* that the age is same. So, the probability we are looking for is 1/200 = 0,5% * Excel calulated it to be 1 in 200.2 times. I simplified it to make the answer look elegant 0 Share this post Link to post Share on other sites

0 Posted 22 Jun 2014 The youngest child among the oldest children in each column is NEVER younger than the oldest child among the youngest children in each row. He is either older or of the same age. The same age happens when the youngest in any row is the oldest in that column. Number of ways in which the ages could be same can be calculated as below: Choose any place (chosen cell) in the 5x10 matrix. This can be done in 50 ways The kid in this cell must be the youngest in the row and oldest in the column. All other places can be filled randomly. So, there are 9 (in row) and 4(in column) "special cells" where combinations must be made. For other cells, the rest can be placed in 36! Ways. The 9 chosen kids in rows and 4 chosen in column can be palced in 9!*4! Ways. Now the only thing that remains to do is to choose 1 kid for the chosen cell and 9 kids for the rows and 4 kids for the columns in the special cells. For any given kid lets say there are x kids younger than him/her and 49-x elder kids Now we need to choose 4 kids out of x younger kids and 9 kids out of 49-x elder kids. Note that x varies from 4 to 40 ONLY. ^{4}C_{4} * ^{45}C_{9} + ^{5}C_{4} * ^{44}C_{9} + .... + ^{40}C_{4} * ^{9}C_{9} Using excel for this, the total combinations comes out to 9.378E + 11 Now, the total possible ways of placing the kids is 50! Total ways in which the age is same is = 50 * 36! * 4! * 9! * the above combinations Relying on excel getting the calculations right, there is 1 in 200 times* that the age is same. So, the probability we are looking for is 1/200 = 0,5% * Excel calulated it to be 1 in 200.2 times. I simplified it to make the answer look elegant Nicely done 0 Share this post Link to post Share on other sites

Posted

Fifty school children lined up at random in five rows and ten columns.

No two of the children have the same birthday.

What is the probability that

the youngest child among the oldest children in each column is

NOT older than

the oldest child among the youngest children in each row ?

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