IF `""^10P_r=604800 and "^10C_r=120` then r=

Options

A. 5
B. 6
C. 7
D. none of these

Answer

To solve the problem, we need to find the value of \( r \) given the equations for permutations and combinations. We have:

1. \( ^{10}P_r = 604800 \)
2. \( ^{10}C_r = 120 \)

### Step 1: Write the equations for permutations and combinations.

The formula for permutations is:
\[ ^{n}P_r = \frac{n!}{(n-r)!} \]
For our case:
\[ ^{10}P_r = \frac{10!}{(10-r)!} = 604800 \tag{1} \]

The formula for combinations is:
\[ ^{n}C_r = \frac{n!}{r!(n-r)!} \]
For our case:
\[ ^{10}C_r = \frac{10!}{r!(10-r)!} = 120 \tag{2} \]

### Step 2: Divide equation (1) by equation (2).

We can simplify the problem by dividing the permutation equation by the combination equation:
\[ \frac{^{10}P_r}{^{10}C_r} = \frac{\frac{10!}{(10-r)!}}{\frac{10!}{r!(10-r)!}} = \frac{r!}{1} \]

So we have:
\[ \frac{604800}{120} = r! \tag{3} \]

### Step 3: Calculate \( r! \).

Now we can compute the left-hand side:
\[ \frac{604800}{120} = 5040 \]
Thus, we have:
\[ r! = 5040 \tag{4} \]

### Step 4: Find \( r \).

Now we need to find the value of \( r \) such that \( r! = 5040 \). We can calculate the factorials of integers to find \( r \):
- \( 1! = 1 \)
- \( 2! = 2 \)
- \( 3! = 6 \)
- \( 4! = 24 \)
- \( 5! = 120 \)
- \( 6! = 720 \)
- \( 7! = 5040 \)

From this, we see that:
\[ r = 7 \]

### Final Answer:
Thus, the value of \( r \) is \( \boxed{7} \).

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