A father with 8 children takes 3 at a time to the Zoological Gardens, as often as he can without taking the same 3 children together more than once. The number of times he will go to the garden, is

Options

A. 336
B. 112
C. 56
D. none of these

Answer

To solve the problem of how many times the father can take 3 out of his 8 children to the Zoological Gardens without repeating the same group of 3, we can use the concept of combinations. Here’s a step-by-step solution:

### Step 1: Understand the Problem
The father has 8 children and he wants to take them to the garden in groups of 3. We need to find out how many unique groups of 3 can be formed from these 8 children.

### Step 2: Use the Combinations Formula
The number of ways to choose r objects from a set of n objects is given by the formula:
\[ nCr = \frac{n!}{r!(n-r)!} \]
In this case, \( n = 8 \) (the total number of children) and \( r = 3 \) (the number of children to take at a time).

### Step 3: Substitute the Values into the Formula
Using the formula, we can calculate \( 8C3 \):
\[ 8C3 = \frac{8!}{3!(8-3)!} = \frac{8!}{3! \cdot 5!} \]

### Step 4: Simplify the Factorials
We can simplify \( 8! \) as follows:
\[ 8! = 8 \times 7 \times 6 \times 5! \]
This allows us to cancel out \( 5! \) in the numerator and denominator:
\[ 8C3 = \frac{8 \times 7 \times 6 \times 5!}{3! \cdot 5!} = \frac{8 \times 7 \times 6}{3!} \]

### Step 5: Calculate \( 3! \)
Now, calculate \( 3! \):
\[ 3! = 3 \times 2 \times 1 = 6 \]

### Step 6: Substitute \( 3! \) Back into the Equation
Now we can substitute \( 3! \) back into our equation:
\[ 8C3 = \frac{8 \times 7 \times 6}{6} \]

### Step 7: Simplify the Expression
Now, simplify the expression:
\[ 8C3 = 8 \times 7 = 56 \]

### Conclusion
Thus, the number of times the father can take 3 children to the garden without repeating the same group is \( 56 \).

### Final Answer
The number of times he will go to the garden is **56**.

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