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A class consists of 5 boys and 4 girls. Given that one kid can only ho

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A class consists of 5 boys and 4 girls. Given that one kid can only ho  [#permalink]

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New post 06 Jul 2015, 10:31
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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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Re: A class consists of 5 boys and 4 girls. Given that one kid can only ho  [#permalink]

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New post 14 Jul 2015, 14:07
(5c2 * 2!) + (4c2 * 2!)
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Re: A class consists of 5 boys and 4 girls. Given that one kid can only ho  [#permalink]

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New post 14 Jul 2015, 18:20
Number of ways of choosing 2 boys out of 5 boys = 5C2.
Number of ways of choosing 2 girs out of 4 girls = 4C2.

The chosen boy/girl can hold any of the two respective titles, so total number of ways = (5C2 * 2!) + (4C2 * 2!) = 20 + 12 = 32 ways. Ans (C).
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Re: A class consists of 5 boys and 4 girls. Given that one kid can only ho  [#permalink]

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New post 14 Jul 2015, 18:34
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reto 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


Easiest way is to treat it like an arrangements question in the following manner:

From the boys we need to select 2 to be clown and pet: This can be done in 5*4 ways

Similarly for the girls, we have 4*3 ways.

Thus total = 20+12 = 32 ways. Thus C is the correct answer.
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Re: A class consists of 5 boys and 4 girls. Given that one kid can only ho  [#permalink]

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New post 14 Jul 2015, 22:48
hey can somebody please explain when to add combinations and when to mulitply combinations?
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A class consists of 5 boys and 4 girls. Given that one kid can only ho  [#permalink]

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New post 14 Jul 2015, 22:51
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Hi jayanthjanardhan,

the combinations are to be multiplied when there is an AND (intersection of two sets/events) and added when there is an OR (union).
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Re: A class consists of 5 boys and 4 girls. Given that one kid can only ho  [#permalink]

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New post 15 Jul 2015, 01:06
i get it now!...thanks FireStorm!
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Re: A class consists of 5 boys and 4 girls. Given that one kid can only ho  [#permalink]

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New post 28 Sep 2015, 04:45
Why does the order matter?
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Re: A class consists of 5 boys and 4 girls. Given that one kid can only ho  [#permalink]

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New post 28 Sep 2015, 04:49
dor1209 wrote:
Why does the order matter?


Because you have 2 distinct titles to give.

Let's say you have 2 kids A and B and you need to award the 2 different titles to them.

If the order didn't matter, you'll say 1 possible combination but consider the following 2 different cases:

A becoming the clown with B becoming the pet is different from

A becoming the pet and B becoming the clown.

This is the reason why order matters.
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Re: A class consists of 5 boys and 4 girls. Given that one kid can only ho  [#permalink]

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New post 06 Jan 2016, 09:47
good one, forgot about multiplying two and ended up with 16..
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Re: A class consists of 5 boys and 4 girls. Given that one kid can only ho  [#permalink]

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New post 07 Mar 2017, 03:53
5C1*4C2+4C1*3C1=32 ways
Answer C.
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Re: A class consists of 5 boys and 4 girls. Given that one kid can only ho  [#permalink]

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New post 07 Mar 2017, 08:37
1 boy can be chosen for class clown in 5C1 ways. Now there are 4 boys left and Teacher's pet can be chosen from this in 4C1 ways.
Since both these conditions must be fulfilled simultaneously so we have 5C1 X 4C1 as the no. of ways of choosing boys while fulfilling the given conditions.

1 girl can be chosen to be the most beautiful girl in class in 4C1 ways. Now there are 3 girls left and smartest kid on the block can be chosen from this in 3C1 ways.
Since both these conditions must be fulfilled simultaneously so we have 4C1 X 3C1 as the no. of ways of choosing girls while fulfilling the given conditions.

Since either of the given conditions needs to be fulfilled, the required number of ways will be 5C1X4C1 + 4C1X3C1 which is equal to 32.

Ans- C
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Re: A class consists of 5 boys and 4 girls. Given that one kid can only ho  [#permalink]

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New post 09 Mar 2017, 03:59
FireStorm wrote:
Number of ways of choosing 2 boys out of 5 boys = 5C2.
Number of ways of choosing 2 girs out of 4 girls = 4C2.

The chosen boy/girl can hold any of the two respective titles, so total number of ways = (5C2 * 2!) + (4C2 * 2!) = 20 + 12 = 32 ways. Ans (C).


Can anyone explain the bolded part, why have we multiplied 5C2 and 4C2 by 2!?
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Re: A class consists of 5 boys and 4 girls. Given that one kid can only ho  [#permalink]

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New post 09 Mar 2017, 23:28
There are 2 titles and any of the boy can take each of these titles. So for one title there are two options and any one of these 2 boys can take that spot. Hence we multiply by 2!.

Title 1 - Any one of the two boys can take this spot. No. of options is 2.
Title 2 - The remaining boy takes this spot. No. of options is 1.

Since both these options should happen simultaneously so we multiply. So 2!.

The same for the girls selection as well.

Now had there been 3 titles the same would have been multiplied by 3!. you can work out the details in the same way as explained above.
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Re: A class consists of 5 boys and 4 girls. Given that one kid can only ho  [#permalink]

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New post 15 Mar 2017, 16:21
reto 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


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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Re: A class consists of 5 boys and 4 girls. Given that one kid can only ho  [#permalink]

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New post 18 Aug 2018, 21:54
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?

concept: AND = multiply ; OR = Add (therefore Class clown X Teach. pet +Mst beautiful X smartest kid
5C1 X 4C1 + 4C1 X 3C1 => 32 (answer)
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Re: A class consists of 5 boys and 4 girls. Given that one kid can only ho  [#permalink]

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New post 18 Aug 2018, 22:55
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reto 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



POINTS to note:-
    1) OR - since we re looking at two different cases joined by OR and are not related to each other. so you have to ADD the ways of each case
    2) two different titles - so ORDER matters


ways to pick 2 boys to be the class clown and the teacher's pet - 5P2=5*4=20 or
2 girls to be the most beautiful girl in class and the smartest kid on the block - 4P2 = 4*3 = 12

total = 20+12= 32

C
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: A class consists of 5 boys and 4 girls. Given that one kid can only ho  [#permalink]

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New post 18 Aug 2018, 23:23
Best thing would be to use permutations here instead of combinations, but actually you may use any.
We use permutations when we are concerned with the order and combination when we are not.
Here order matters, thus we can use permutation to get a straight answer, however, we may use combinations as well and then multiply by 2! to compensate for the inner arrangement.
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Re: A class consists of 5 boys and 4 girls. Given that one kid can only ho  [#permalink]

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New post 18 Aug 2018, 23:28
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Though i got the answer is as 32.

I suppose if the question had an AND instead of OR.

The answer would be 20 ×12=240

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Re: A class consists of 5 boys and 4 girls. Given that one kid can only ho  [#permalink]

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New post 18 Aug 2018, 23:42
reto 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


Since it is given one kid can hold only one title, it depicts that order of selection is fixed. When order is fixed we perform permutation and not combination.
2 boys as class clown and teacher's pet= 5*4
2 girls as the most beautiful and smartest= 4*3
It is given by OR condition, hence 5*4+4*3=32
Ans: C
Surely not 700 level question
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Re: A class consists of 5 boys and 4 girls. Given that one kid can only ho &nbs [#permalink] 18 Aug 2018, 23:42

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