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

Question

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

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

madgmat2019 · Answer

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

eakabuah · Answer

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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Archit3110 · Answer

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

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

exc4libur · Answer

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

Answer (C)

unraveled · Answer

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

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).

Mohammadmo · Answer

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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lacktutor · Answer

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.

Regor60 · Answer

The disallowed cases are those where neither parent goes, in which case the tickets are allocated solely to the children:

7!/4!3! = 35

These cases must be subtracted from the total number of ways to allocate the tickets among the nine family members:

9!/4!5! = 126

126-35 = 91

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bumpbot · Answer

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

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