Answer:

Step-by-step explanation:
<u>The full question:</u>
<em>"A committee has eleven members. there are 3 members that currently serve as the boards chairman, ranking members, and treasurer. each member is equally likely to serve in any of the positions. Three members are randomly selected and assigned to be the new chairman, ranking member, and treasurer. What is the probability of randomly selecting the three members who currently hold the positions of chairman, ranking member, and treasurer and reassigning them to their current positions?"</em>
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The permutation of choosing 3 members from a group of 11 would be:
P(n,r) = 
Where n would be the total [in this case n is 11] & r would be 3
Which is:
P(11,3) = 
So there are total of 990 possible way and there is ONLY ONE WAY for them to be reassigned. Hence the probability would be:
1/990
Answer:
x = 5 y = 9
Step-by-step explanation:
74 R28
So sorry if its incorrect!
Answer:
Adult ticket: $7
Child ticket: $2
Step-by-step explanation:
Set up a system of equations where a represents the cost of one adult ticket and c is the cost of one child ticket:
2a + 3c = 20
a + 4c = 15
Solve by elimination by multiplying the bottom equation by -2:
2a + 3c = 20
-2a -8c = -30
Add them together:
-5c = -10
c = 2
Now, we can plug in 2 as c to find the value of a:
2a + 3c = 20
2a + 3(2) = 20
2a + 6 = 20
2a = 14
a = 7
The number of permutations of the 4 different digits is:

The answer is 24 ways.