T1 = 3
t2 = 2*t1 = 6
t3 = 2*t2 = 12
t4 = 24
t5 = 48
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
180a^3+900a^3+4a-5
__________________ + a-1/ 6a^a
6a ( a+ 5)
Answer:
Step-by-step explanation:
There are choices when it comes to color (3). Then there is the choice between a boy dog or a girl dog (2). Multiply the number of colors by the number of genders. (6)
When choosing a dog from random, you can either choose:
Brown/M
Brown/F
Black/M
Black/F
Yellow/M
Yellow/F
(This list is called the Sample Space, a list of all possible outcomes)
Because the text says that all types of dogs are found in even numbers, it would not matter if there was 6, 12, 18, ... amount of dogs.
When looking at this sample space, you can see that the chance of choosing a Yellow/F dog, is 1 in 6.