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A mother and father have seven children, and the family receives an in

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A mother and father have seven children, and the family receives an in  [#permalink]

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New post 16 Oct 2016, 11:40
2
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A
B
C
D
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Difficulty:

  75% (hard)

Question Stats:

53% (02:11) correct 47% (01:03) wrong based on 200 sessions

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A mother and father have seven children, and the family receives an invitation with four tickets to the circus. If the family decides to randomly select the four members who get to attend the circus, but determine that the chosen group must include at least one of the mother and father, how many different groups are possible to send to the circus?

A. 70
B. 84
C. 91
D. 108
E. 126
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Re: A mother and father have seven children, and the family receives an in  [#permalink]

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New post 16 Oct 2016, 12:04
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sameerspice wrote:
A mother and father have seven children, and the family receives an invitation with four tickets to the circus. If the family decides to randomly select the four members who get to attend the circus, but determine that the chosen group must include at least one of the mother and father, how many different groups are possible to send to the circus?

A. 70
B. 84
C. 91
D. 108
E. 126


Condition 1: Atleast one of the mother and father along with seven children must be selected

Then either - Mother or father = 2C1 = 2 ways for one ticket and 7C3 = 35 ways for other three tickets. Total ways here is 2*35 = 70

Condition 2: Both M and F selected and this is 1 way...then we'll have only two tickets for 7 children..then we can select 7C2 = 21.

Total 70+21 = 91.

Option C.
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A mother and father have seven children, and the family receives an in  [#permalink]

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New post 19 Jan 2017, 01:27
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I think this is a shorter solution.
All possible selections = 9C4 (Selecting any 4 members out of the total 9 members).
Selections involving only children (i.e. no parent is selected) = 7C4(Selecting 4 children out of 7 children).
Now 9C4 - 7C4 = Scenarios where atleast one parent has to be selected = 126-35 = 91

Hence C. :)
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Re: A mother and father have seven children, and the family receives an in  [#permalink]

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New post 20 Jan 2017, 09:21
I made this approach, which i saw it in similar examples but there is something wrong.
For the first seat we have 2 options (father or mother) the second one 8 options (7children remaining parent) third 7 options 4th 6 options
So 2*8*7*4/ 4! = 28
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Re: A mother and father have seven children, and the family receives an in  [#permalink]

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New post 22 Jan 2017, 00:00
Solve these questions systematically. We can have 3 cases. Case 1: When the father goes to the circus along with his 3 children. So F,C1,C2,C3. This can happen in 7C3 ways. Case 2: When the mother goes accompanied by her 3 children. So M,C1,C2,C3. This can happen in 7C3 ways. Finally, Case 3: When both the parents go along with 2 children. So, M,F,C1,C2. This can happen in 7C2 ways. So the total number of possible combinations= 2*7C3+7C2= 70+21=91.
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A mother and father have seven children, and the family receives an in  [#permalink]

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New post 25 Sep 2018, 03:00
2 Ways to do this:

Quote:
Because the parents are the limiting factor here - at least one must be in the final group - it can be helpful to begin by thinking through the situations that will work: Mom goes and Dad doesn't; Dad goes and Mom doesn't; and both parents go (this is the one people forget!). With each of those situations:

Mom goes, Dad doesn't: that leaves 3 spots left to select from the 7 children. Using the combinations formula that gives 7!/3!4!=7∗6∗5/3∗2=35

Dad goes, Mom doesn't: that math is the same as above, with 7 children vying for the 3 remaining spots. Using the combinations formula that's: 7!/3!4!=7∗6∗5/3∗2=35

Both parents go: that leaves the 7 children with only 2 spots to be chosen. The combinations formula then looks like: 7!/2!5!=7∗6/2=21

So the three different possible situations have 35, 35, and 21 outcomes each. The sum of those is 91 total options that will work.


Quote:
Alternatively, you could start with the total number of unrestricted groups (9 family members, choose 4) and work backward by subtracting those that violate the constraint (those groups that consist of 4 children with no parents). Were there no restriction and there were 9 people vying for 4 spots, that math would be:

9!/4!5!=9∗8∗7∗6/4∗3∗2∗1=126

But then you must subtract the "no parents / all children" groups. To find those, take the parents out of the pool of candidates, meaning that there are 7 people from which you'll select 4. That math is 7!/3!4!=7∗6∗5/3∗2=35

Therefore, 126 - 35 = 91, again leading to answer choice C.

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