on bingo boards each column has numbers that come from a range of values. For example the first column could be 0-19 the second 20-39 etc (20 possibilities per column)
If 5 numbers are chosen at random to fill the first column only how many ways can the first column be filled?
2006-08-25 13:19:13 · 5 answers · asked by Tony M 1 in Education & Reference ➔ Higher Education (University +)
One possible answer is 45,326.
But there isn't enough information to answer the question. Bingo uses the numbers 1 to 75 (US) or 1 to 90 (non US). If the columns of your bingo boards have a range of 20 numbers in them then your bingo board has either 3¾ or 4½ columns in it.
Alternatively, assuming your bingo board has a range of 20 numbers per column then the number of bingo balls you have should be 20 multipled by the number of rows on your bingo board.
The calculation can only be solved by knowing the total number of bingo balls.
However, if it were a standard US bingo card which has 5 columns the first of which contains 5 numbers in the range 1 to 15, the second contains 5 numbers in the range 16 to 30, 3rd column range is 31 to 45, 4th column = 46 to 60, 5th column = 61 to 75.
15 possible numbers in the first column. Subsequent rows are always higher numbers. First column has to be a number from 1 to 11, second column 2 to 12, 3rd column 3 to 13, 4th column 4 to 14, 5th column 5 to 15.
There are 11 possible numbers in the first column.
There are 11 possible numbers in the second column but... you need to take into account that any one of 10 of those numbers may have been used in cloumn 1
There are 11 possible numbers in the third column but... you need to take into account that any one of 9 of those numbers may have been used in cloumn 1 and that any one of 10 of those numbers may have been used in cloumn 2
There are 11 possible numbers in the fourth column but... you need to take into account that any one of 8 of those numbers may have been used in cloumn 1 and that any one of 9 of those numbers may have been used in cloumn 2 and that any one of 8 of those numbers may have been used in cloumn 3
There are 11 possible numbers in the fifth column but... you need to take into account that any one of 7 of those numbers may have been used in cloumn 1 and that any one of 8 of those numbers may have been used in cloumn 2 and that any one of 9 of those numbers may have been used in cloumn 3 and that any one of 10 of those numbers may have been used in cloumn 4
The equation for solving it is therefore...
11 + ((10 x (10/11))+1) + ((9 x (9/11) x (10/11))+2) + ((8 x (8/11) x (9/11) x (10/11))+3) + ((7 x (7/11) x (8/11) x (9/11) x (10/11)) +4) = 45,326
2006-08-25 14:27:46 · answer #1 · answered by Trevor 7 · 0⤊ 0⤋
There are 20 choices for the first number.
Therefore, there are only 19 choices for the 2nd.
Then 18 and so on.
The answer is, if you are choosing 5 random number out of 20, there are (20 * 19 * 18 * 17 * 16) different combinations of numbers.
2006-08-25 20:38:05 · answer #2 · answered by sassy_91 4 · 0⤊ 0⤋
75 ways.Without using any of the values twice.Because there are 15 values in the first column of bingo. B1-B15
2006-08-25 21:48:05 · answer #3 · answered by Purple Passion 3 · 0⤊ 0⤋
Who cares your cheating on your final Buddy Boy! While the teachers out playing ball! lol..... guess who
2006-08-25 20:53:57 · answer #4 · answered by Sad Mom 3 · 0⤊ 0⤋
i have no idea
2006-08-25 20:25:03 · answer #5 · answered by danzar279 3 · 0⤊ 0⤋