The answer is 12C5 = 12*11*10*9*8/(5*4*3*2*1) = 792
The number of ways is 18C5 = 18!/(5!*13!) = 8,568 ways.
Since each ticket wins, the question is equivalent to the number of ways of selecting 4 tickets out of 20. This is 4845.
30 = 6 * 5 if we assume the president and the vice-president must be different people.
There are 8!/(4!*4!) = 70 ways.
Three students can be selected from 5 in (5 x 4 x 3) = 60 ways.BUT there are (3 x 2) = 6 ways to select the same 3 students.So there are only 60/6 = 10 different groups of 3 studentsthat can be selected from a pool of 5.
53,130 ways.
There are 230300 ways.
435 ways.
The number of ways is 18C5 = 18!/(5!*13!) = 8,568 ways.
Q9. How many different ways can 4 tickets be selected from 50 tickets if each ticket wins a different prize.
In how many ways can fourfour women be selected from the eighteight ​women
i think about 5929 ways
They can be selected in 756 ways.
I would say 77 because if there are 77 books, you could put each different book in a different place 77 times.
Since each ticket wins, the question is equivalent to the number of ways of selecting 4 tickets out of 20. This is 4845.
If there are 58 defective circuit boards, two can be selected in 58*57/2 = 1653 ways.
18x17= 306 ways