Question 1042072
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Use the combination formula to get


n C r = (n!)/(r!(n-r)!)


7 C 5 = (7!)/(5!*(7-5)!)


7 C 5 = (7!)/(5!*2!)


7 C 5 = (7*6*5!)/(5!*2!)


7 C 5 = (7*6)/(2!)


7 C 5 = (7*6)/(2*1)


7 C 5 = (42)/(2)


7 C 5 = 21


So there are 21 ways to pick 5 students. The sample space would be...


A,B,C,D,E
A,B,C,D,F
A,B,C,D,G
A,B,C,E,F
A,B,C,E,G
A,B,C,F,G
A,B,D,E,F
A,B,D,E,G
A,B,D,F,G
A,B,E,F,G
A,C,D,E,F
A,C,D,E,G
A,C,D,F,G
A,C,E,F,G
A,D,E,F,G
B,C,D,E,F
B,C,D,E,G
B,C,D,F,G
B,C,E,F,G
B,D,E,F,G
C,D,E,F,G


If you counted all the possibilities in the sample space above, you'd count out 21 different items. Each item has its own line.


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Now let's lock up the first two slots for students A and B. That leaves 7-2 = 5 students left and 5-2 = 3 slots left


So using the combination formula again gives


n C r = (n!)/(r!(n-r)!)


5 C 3 = (5!)/(3!*(5-3)!)


5 C 3 = (5!)/(3!*2!)


5 C 3 = (5*4*3!)/(3!*2!)


5 C 3 = (5*4)/(2!)


5 C 3 = (5*4)/(2*1)


5 C 3 = (20)/(2)


5 C 3 = 10


There are 10 ways to pick a group of five students (from the pool of seven) where students A and B are guaranteed to go. 


The 10 different ways are listed below


A,B,C,D,E
A,B,C,D,F
A,B,C,D,G
A,B,C,E,F
A,B,C,E,G
A,B,C,F,G
A,B,D,E,F
A,B,D,E,G
A,B,D,F,G
A,B,E,F,G


That list of 10 is a subset of the first list of items shown above. 


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Now divide the two values to get <font color=red size=4>10/21</font> which is the final answer



Side note: 10/21 = 0.47619047619048 = 47.619047619048%


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Final Answer: <font color=red size=4>10/21</font>
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