A fruit basket contains 6 oranges, 8 apples and 12 mangoes. In how many ways is it possible to make a selection by taking any number of 15 fruits, namely 3 oranges, 5 apples and 7 mangoes?

Options

A. 887040
B. 887400
C. 878500
D. 887060

Answer

To solve the problem of selecting fruits from the basket, we need to calculate the number of ways to choose specific quantities of oranges, apples, and mangoes from the available amounts. Here’s a step-by-step breakdown:

### Step 1: Identify the quantities
We have:
- 6 oranges, and we need to select 3.
- 8 apples, and we need to select 5.
- 12 mangoes, and we need to select 7.

### Step 2: Set up the combinations
We will use the combination formula \( nCk \), which is given by:
\[ nCk = \frac{n!}{k!(n-k)!} \]
where \( n \) is the total number of items to choose from, and \( k \) is the number of items to choose.

### Step 3: Calculate the combinations for each fruit
1. **For oranges**: We need to calculate \( 6C3 \):
\[ 6C3 = \frac{6!}{3!(6-3)!} = \frac{6!}{3! \times 3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]

2. **For apples**: We need to calculate \( 8C5 \):
\[ 8C5 = \frac{8!}{5!(8-5)!} = \frac{8!}{5! \times 3!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \]

3. **For mangoes**: We need to calculate \( 12C7 \):
\[ 12C7 = \frac{12!}{7!(12-7)!} = \frac{12!}{7! \times 5!} = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} = 792 \]

### Step 4: Calculate the total number of ways
Now, we multiply the number of ways to choose each type of fruit:
\[ \text{Total ways} = 6C3 \times 8C5 \times 12C7 = 20 \times 56 \times 792 \]

### Step 5: Perform the multiplication
1. First, calculate \( 20 \times 56 \):
\[ 20 \times 56 = 1120 \]

2. Now, multiply \( 1120 \times 792 \):
\[ 1120 \times 792 = 886560 \]

### Conclusion
The total number of ways to select 3 oranges, 5 apples, and 7 mangoes from the basket is **886560**.