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Bunuel
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A class consists of 5 boys and 4 girls. Given that one kid can only ho 25 Nov 2019, 01:22
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60% (01:31) correct 40% (01:34) wrong based on 359 sessions

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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


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C
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A class consists of 5 boys and 4 girls. Given that one kid can only ho 25 Nov 2019, 03:30
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for boyes = 5c2*2! ; 20
for girls = 4c2*2! ; 12
total 32
IMO C


Bunuel
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


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A class consists of 5 boys and 4 girls. Given that one kid can only ho 25 Nov 2019, 09:49
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Bunuel
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
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5C2 (ways to choose 2 titles of 5 boys) * 2 (pick clown first pet second, or pet first clown second) = 20
4C2 … * 2 … = 12
Total = 20+12 = 32

Ans (C)
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A class consists of 5 boys and 4 girls. Given that one kid can only ho 11 Jun 2021, 03:35
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By FCP

For Boys, All boys stand a chance to have a titles so 5 ways and then there are only 4 other boys that will remain for the other title so total ways will be 5*4=20
Same process for girls 4 ways and 3 ways so 3*4=12
Total ways= addition of boys ways and girls ways because of the word OR
So 20+12=32ways
Option C is the ans
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A class consists of 5 boys and 4 girls. Given that one kid can only ho 01 Dec 2025, 22:02
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Bunuel KarishmaB Kinshook Archit3110 nick1816

Shouldn't the answer here be 64? Which of course is not one of the options here :(

No. of ways to choose class clown and pet among 5 boys = 5C1 X 4C1 X 2! = 40 (2! since order here matters)

No. of ways to choose beautiful and smart among 4 girls = 4C1 X 3C1 X 2! = 24 (2! since order matter again)

Therefore, total no. of ways = 40 + 24 = 64

Bunuel
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


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A class consists of 5 boys and 4 girls. Given that one kid can only ho 01 Dec 2025, 23:40
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Faszbendarr
Bunuel KarishmaB Kinshook Archit3110 nick1816

Shouldn't the answer here be 64? Which of course is not one of the options here :(

No. of ways to choose class clown and pet among 5 boys = 5C1 X 4C1 X 2! = 40 (2! since order here matters)

No. of ways to choose beautiful and smart among 4 girls = 4C1 X 3C1 X 2! = 24 (2! since order matter again)

Therefore, total no. of ways = 40 + 24 = 64


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Your extra 2! is the issue.

When you do 5C1 * 4C1, you are already assigning different roles: first choice = class clown, second choice = teacher's pet. That already takes order into account, so it equals 5P2 = 20. Multiplying by 2! again double counts each pair.

Same for the girls part.
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A class consists of 5 boys and 4 girls. Given that one kid can only ho 24 Jan 2026, 13:15
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Can someone explain why we're adding 20 and 12 instead of multiplying for 240?
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A class consists of 5 boys and 4 girls. Given that one kid can only ho 25 Jan 2026, 11:16
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roshaun25
Can someone explain why we're adding 20 and 12 instead of multiplying for 240?
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The question asks "...in how many ways can you pick 2 boys to be .... OR 2 girls to be ...."

If it were AND we would have multiplied, because we calculate for cases where both exist together. Since, it is OR we add, because it is "either, or" and multiplication leads to cases where both exist together.
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