To solve the problem of finding the number of shortest ways to reach from the point (0, 0, 0) to the point (3, 7, 11) in space, we can follow these steps:
### Step 1: Determine the total distance
To reach from (0, 0, 0) to (3, 7, 11), we need to move:
- 3 units along the x-axis,
- 7 units along the y-axis,
- 11 units along the z-axis.
The total distance (or total number of moves) is:
\[
3 + 7 + 11 = 21
\]
### Step 2: Calculate the number of ways to arrange the moves
The total number of moves is 21, and we need to choose how to arrange these moves. We can think of this as arranging a sequence of moves where:
- 'X' represents a move along the x-axis,
- 'Y' represents a move along the y-axis,
- 'Z' represents a move along the z-axis.
Thus, we have:
- 3 moves in the x-direction (X),
- 7 moves in the y-direction (Y),
- 11 moves in the z-direction (Z).
The total number of arrangements of these moves can be calculated using the multinomial coefficient:
\[
\text{Number of ways} = \frac{21!}{3! \times 7! \times 11!}
\]
### Step 3: Calculate the factorial values
Now, we will compute the factorial values:
- \( 21! \) is the factorial of 21,
- \( 3! = 6 \),
- \( 7! = 5040 \),
- \( 11! = 39916800 \).
### Step 4: Substitute the values into the formula
Substituting these values into the formula gives:
\[
\text{Number of ways} = \frac{21!}{3! \times 7! \times 11!} = \frac{21!}{6 \times 5040 \times 39916800}
\]
### Step 5: Calculate the final result
Now, we can compute \( 21! \) and then divide by the product of \( 3! \), \( 7! \), and \( 11! \):
1. Calculate \( 21! \) (which is a large number),
2. Calculate the denominator \( 6 \times 5040 \times 39916800 \),
3. Finally, perform the division to find the total number of shortest paths.
### Final Answer
After performing the calculations, we find that the total number of shortest ways to reach from (0, 0, 0) to (3, 7, 11) is \( 116280 \).
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