The number of times will an individual child goes to the cinema before a group is repeated.

Question

$1.)$ A mother with $7$ children takes $3$ at a time to a cinema.She goes with every group of $3$ that she can form.How many times can she go to cinema with distinct groups of $3$ children?

$2.)$ For the above question , how many times will an individual child go to the cinema with her before a group is repeated?

For the $1$st question I think the answer is $\dbinom{7}{3}$

for the $2$nd one , I am simply unable to interpret the question though the answer given for question $2.)$ is $\dbinom{6}{2}$.

I look for a short and simple way.

I have studied maths upto $12$th grade.

Answer 1

Part $1$ is correct.

Part $2$ is also correct. It can also be written and solved as the following:

$$\sum_{i=1}^5i$$

Nice job!

Answer 2

1st one is ok.

For the 2nd one, either consider that you must be attached to all groups of 2 from the remaining, hence ${6\choose2}$, or that you can only be in $\frac37$ of the 35 possible groups.

Answer 3

There are ${7\choose 3}=35$ different groups of three that can be formed. Realizing all of them means $35$ trips to the cinema, or $35\cdot3=105$ children's tickets. Since there are $7$ children each of them has visited the cinema ${105\over7}=15={6\choose 2}$ times, namely with every pair of his $6$ siblings.