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

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Joined: 02 Sep 2009
Posts: 58332
A mother and father have seven children, and the family receives an in  [#permalink]

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23 Sep 2019, 00:01
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Difficulty:

55% (hard)

Question Stats:

61% (01:51) correct 39% (01:08) wrong based on 43 sessions

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Competition Mode Question

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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23 Sep 2019, 00:14
1
two possiblities are (one of the two parents*3 out of 7 children + (2 out of 2 parents * 2 out of 7 children)
=(2c1*7c3 + 2c2*7c2) = 91
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Re: A mother and father have seven children, and the family receives an in  [#permalink]

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23 Sep 2019, 00:19
Number of choosing either mother or father = 2C1=2
Number of ways of choosing 3 out of 7 children = 7C3 = 35
Total number of choosing a parent and 3 children = 2*35 = 70

Answer is A.

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

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23 Sep 2019, 00:26
1
Mother and father be 1 set and rest 7 be another set ; so for 4 tickets with condition chosen group must include at least one of the mother and father.
2c1*7c3+2c2*7c2 = 70+21 =91
IMO C

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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Posts: 576
Location: United States
Re: A mother and father have seven children, and the family receives an in  [#permalink]

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23 Sep 2019, 03:13
1
Quote:
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

[1] one parent and three children: 2C1•7C3=70
[2] two parents and two children: 2C2•7C2=21
[1 or 2]: 70+21=91

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

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23 Sep 2019, 03:35
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

Since only four tickets are available out of which one would be either a mother or a father in 2C1 ways, the remaining three would be filled among seven in 7C3 ways.
Thus, total possible ways = 2C1 * 7C3
= 70

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

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23 Sep 2019, 03:54
We have two ways to select
1)Mother or father along with 3 of their children.2C1*7C3=2×35=70
2)both mother and father along with 2 of their children.2C2*7C2= 1×21.
So,70+21=91
Hence option C

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

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23 Sep 2019, 16:47
4 invitations to the circus FOR:
--> A mother and a father
--> and 7 kids

At least one of a mother or a father must include in the group of FOUR people.
---How many different groups are possible to send to the circus?

There are two cases for the solution:

1) if one of a mother or a father goes to the circus, then
--> 2C1 * 7C3= 2*$$\frac{7!}{3!*4!}$$=2*$$\frac{7*6*5}{3*2*1}$$=2*35=70

2) If Both of them go to the circus, then
--> 2C2 * 7C2=$$\frac{7!}{2!*5!}$$=$$\frac{7*6}{2}$$=21

The total different groups are 70+21=91

The answer is C.
Re: A mother and father have seven children, and the family receives an in   [#permalink] 23 Sep 2019, 16:47
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