7) A theater is staging a series of 10 different plays.You want to attend at least 2 of the plays.How many different combinations of plays can you attend?
10C2 = 10!/((10-2)!2!)
x!=factorial x=x(x-1)(x-2)....(2)(1)
To find out the number of different combinations of plays you can attend, you can use the concept of combinations. In this case, you want to attend at least 2 plays out of a series of 10 different plays.
To calculate the number of combinations, you can use the formula for combinations:
C(n, r) = n! / (r! * (n - r)!)
In this formula, "n" represents the total number of items (in this case, plays), and "r" represents the number of items you want to choose (in this case, 2 or more plays).
For your scenario, you need to calculate the combinations for attending 2, 3, 4, 5, 6, 7, 8, 9, and 10 plays, as you want to attend at least 2 plays.
C(10, 2) = 10! / (2! * (10 - 2)!)
C(10, 3) = 10! / (3! * (10 - 3)!)
C(10, 4) = 10! / (4! * (10 - 4)!)
C(10, 5) = 10! / (5! * (10 - 5)!)
C(10, 6) = 10! / (6! * (10 - 6)!)
C(10, 7) = 10! / (7! * (10 - 7)!)
C(10, 8) = 10! / (8! * (10 - 8)!)
C(10, 9) = 10! / (9! * (10 - 9)!)
C(10, 10) = 10! / (10! * (10 - 10)!)
Now, let's calculate these combinations:
C(10, 2) = 10! / (2! * 8!)
= (10 * 9) / (2 * 1)
= 45
C(10, 3) = 10! / (3! * 7!)
= (10 * 9 * 8) / (3 * 2 * 1)
= 120
C(10, 4) = 10! / (4! * 6!)
= (10 * 9 * 8 * 7) / (4 * 3 * 2 * 1)
= 210
C(10, 5) = 10! / (5! * 5!)
= (10 * 9 * 8 * 7 * 6) / (5 * 4 * 3 * 2 * 1)
= 252
C(10, 6) = 10! / (6! * 4!)
= (10 * 9 * 8 * 7 * 6 * 5) / (6 * 5 * 4 * 3 * 2 * 1)
= 210
C(10, 7) = 10! / (7! * 3!)
= (10 * 9 * 8 * 7 * 6 * 5 * 4) / (7 * 6 * 5 * 4 * 3 * 2 * 1)
= 120
C(10, 8) = 10! / (8! * 2!)
= (10 * 9 * 8 * 7 * 6 * 5 * 4 * 3) / (8 * 7 * 6 * 5 * 4 * 3 * 2 * 1)
= 45
C(10, 9) = 10! / (9! * 1!)
= (10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2) / (9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1)
= 10
C(10, 10) = 10! / (10! * 0!)
= (10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1) / (10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1)
= 1
Now, add up all the calculated combinations:
45 + 120 + 210 + 252 + 210 + 120 + 45 + 10 + 1 = 1,053
Therefore, there are 1,053 different combinations of plays that you can attend.