A man has 8children to take them to a zoo. He takes thre of them at a time to the zoo as often as he can without taking the same 3 children together more than once. How many times will he have to go to the zoo? How many times a particular child will to to the zoo?

Options

A. 56
B. 21
C. 112
D. none of these

Answer

To solve the problem step by step, we need to determine two things: how many times the man will go to the zoo with his children and how many times a particular child will go to the zoo.

### Step 1: Determine the total number of ways to choose 3 children from 8
The man has 8 children and he takes 3 of them at a time to the zoo. We can calculate the number of ways to choose 3 children from 8 using the combination formula:

\[ \text{Number of ways} = \binom{n}{r} = \frac{n!}{r!(n-r)!} \]

In this case, \(n = 8\) and \(r = 3\):

\[ \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8!}{3! \cdot 5!} \]

### Step 2: Simplify the combination
We can simplify this expression:

\[ \binom{8}{3} = \frac{8 \times 7 \times 6 \times 5!}{3! \times 5!} \]

The \(5!\) in the numerator and denominator cancels out:

\[ = \frac{8 \times 7 \times 6}{3!} \]

Now, calculate \(3!\):

\[ 3! = 3 \times 2 \times 1 = 6 \]

So we have:

\[ \binom{8}{3} = \frac{8 \times 7 \times 6}{6} = 8 \times 7 = 56 \]

### Conclusion for Step 1
The man will have to go to the zoo **56 times**.

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### Step 3: Determine how many times a particular child will go to the zoo
Now, we need to find out how many times a particular child (let's say Child A) will go to the zoo.

If we fix Child A, we need to select 2 more children from the remaining 7 children. The number of ways to choose 2 children from 7 is given by:

\[ \binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7!}{2! \cdot 5!} \]

### Step 4: Simplify this combination
Similar to before, we simplify:

\[ \binom{7}{2} = \frac{7 \times 6 \times 5!}{2! \times 5!} \]

Again, the \(5!\) cancels out:

\[ = \frac{7 \times 6}{2!} \]

Calculating \(2!\):

\[ 2! = 2 \times 1 = 2 \]

So we have:

\[ \binom{7}{2} = \frac{7 \times 6}{2} = \frac{42}{2} = 21 \]

### Conclusion for Step 2
A particular child will go to the zoo **21 times**.

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### Final Answers
- The man will go to the zoo **56 times**.
- A particular child will go to the zoo **21 times**.

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