The number of ways of arranging `6` players to throw the cricket ball so that oldest player may not throw first is

Options

A. `120`
B. `600`
C. `720`
D. `7156`

Explanation

To solve the problem of arranging 6 players to throw a cricket ball such that the oldest player does not throw first, we can follow these steps:

### Step 1: Identify the players
Let's denote the players as P1, P2, P3, P4, P5, and P6, where P6 is the oldest player.

### Step 2: Determine the first thrower
Since the oldest player (P6) cannot throw first, we have 5 options for the first thrower. The possible players for the first throw are P1, P2, P3, P4, and P5.

### Step 3: Arrange the remaining players
After choosing the first thrower, we have 5 players left (including P6) to arrange in the remaining 5 positions. The number of ways to arrange these 5 players is given by the factorial of the number of players, which is 5!.

### Step 4: Calculate the total arrangements
The total number of arrangements can be calculated as follows:
- For the first throw, we have 5 choices (P1, P2, P3, P4, P5).
- For the remaining 5 players, we can arrange them in 5! ways.

Thus, the total number of arrangements is:
\[ \text{Total arrangements} = 5 \times 5! \]

### Step 5: Calculate 5!
Calculating 5!:
\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]

### Step 6: Calculate the final answer
Now substituting back into the total arrangements:
\[ \text{Total arrangements} = 5 \times 120 = 600 \]

### Conclusion
Therefore, the number of ways of arranging the 6 players so that the oldest player does not throw first is **600**.

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