A class consists of 5 boys and 4 girls. Given that one kid can only hold one title, in how many ways can you pick 2 boys to be the class clown and the teacher's pet or 2 girls to be the most beautiful girl in class and the smartest kid on the block?
A. 9
B. 18
C. 32
D. 60
E. 240
The OA is C.
Please, can somebody explain when to make the sum of combinations and when to do the product of combinations? Thanks!
A class consists of 5 boys and 4 girls. Given that one kid
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The class consists of 5 boys and 4 girls.BTGmoderatorLU wrote:A class consists of 5 boys and 4 girls. Given that one kid can only hold one title, in how many ways can you pick 2 boys to be the class clown and the teacher's pet or 2 girls to be the most beautiful girl in class and the smartest kid on the block?
A. 9
B. 18
C. 32
D. 60
E. 240
The OA is C.
Please, can somebody explain when to make the sum of combinations and when to do the product of combinations? Thanks!
Given: One kid can only hold one title
Number of ways a boy can be selected as the class clown = 5;
Number of ways a boy can be selected as the teacher's pet = 4; Since one boy is already selected a clown, the number of boys left = 4
This is a case of AND, so we multiply the number of ways
Number of ways 2 boys of the class can be the class clown AND the teacher's pet = 5*4 = 20 ---(1)
Number of ways a girl can be selected as the most beautiful girl = 4;
Number of ways a girl can be selected as the smartest kid = 3; Since one girl is already selected as the most beautiful girl, the number of girls left = 3
This is a case of AND, so we multiply the number of ways
Number of ways 2 girls of the class can be the most beautiful girl AND the smartest kid = 4*3 = 12 ---(2)
Since the situation is that we have to choose either 2 boys OR 2 girls, we will add the total number of ways [(1) + (2)].
Total number of ways = 20 + 12 = 32
The correct answer: C
Hope this helps!
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First selected two kids out of boys or girls and since the position for two selected kids are different so arrangement of two selected kids may happen in 2! ways.BTGmoderatorLU wrote:A class consists of 5 boys and 4 girls. Given that one kid can only hold one title, in how many ways can you pick 2 boys to be the class clown and the teacher's pet or 2 girls to be the most beautiful girl in class and the smartest kid on the block?
A. 9
B. 18
C. 32
D. 60
E. 240
The OA is C.
Please, can somebody explain when to make the sum of combinations and when to do the product of combinations? Thanks!
If boys take the position allowed for them, The cases = 5C2*2! = 10*2 = 20
If Girls take the position allowed for them, The cases = 4C2*2! = 6*2 = 12
Total Favourable cases = 20+12 = 32
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BTGmoderatorLU wrote:A class consists of 5 boys and 4 girls. Given that one kid can only hold one title, in how many ways can you pick 2 boys to be the class clown and the teacher's pet or 2 girls to be the most beautiful girl in class and the smartest kid on the block?
A. 9
B. 18
C. 32
D. 60
E. 240
The OA is C.
We are given a class of 5 boys and 4 girls.
Let's first determine how many ways 2 boys can be selected for class clown and teacher's pet.
Suppose two of the boys are Adam and Ben, and they are selected as the class clown and the teacher's pet. Saying Adam is the class clown and Ben is the teacher's pet is DIFFERENT from saying Ben is the class clown and Adam is the teacher's pet. Therefore, this is a permutation problem, since the order matters.
Since here we have 5 boys and we are selecting 2 for which order matters, the number of ways this can be done is:
5P2 = 5 x 4 = 20
Similarly, the number of ways 2 girls can be selected from 4 girls to be the most beautiful girl in class and the smartest kid on the block in which order matters is:
4P2 = 4 x 3 = 12
Thus, the number of ways to select them is 20 + 12 = 32.
Answer: C
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