Find the value of 'r' if `""^(5)P_(r )=""^(6)P_(r-1)`.

Answer

To solve the equation \( {}^{5}P_{r} = {}^{6}P_{r-1} \), we will use the formula for permutations, which is given by:

\[ {}^{n}P_{r} = \frac{n!}{(n-r)!} \]

### Step-by-step Solution:

1. **Write the permutation formulas**:
\[ {}^{5}P_{r} = \frac{5!}{(5-r)!} \]
\[ {}^{6}P_{r-1} = \frac{6!}{(6-(r-1))!} = \frac{6!}{(7-r)!} \]

2. **Set the two expressions equal to each other**:
\[ \frac{5!}{(5-r)!} = \frac{6!}{(7-r)!} \]

3. **Substituting the factorials**:
Recall that \( 6! = 6 \times 5! \), so we can rewrite the equation:
\[ \frac{5!}{(5-r)!} = \frac{6 \times 5!}{(7-r)!} \]

4. **Cancel \( 5! \) from both sides** (assuming \( 5! \neq 0 \)):
\[ \frac{1}{(5-r)!} = \frac{6}{(7-r)!} \]

5. **Cross-multiply to eliminate the fractions**:
\[ (7-r)! = 6 \times (5-r)! \]

6. **Expand \( (7-r)! \)**:
\[ (7-r)(6-r)(5-r)! = 6 \times (5-r)! \]

7. **Cancel \( (5-r)! \) from both sides** (assuming \( (5-r)! \neq 0 \)):
\[ (7-r)(6-r) = 6 \]

8. **Expand the left side**:
\[ 42 - 13r + r^2 = 6 \]

9. **Rearranging the equation**:
\[ r^2 - 13r + 36 = 0 \]

10. **Factor the quadratic equation**:
\[ (r - 9)(r - 4) = 0 \]

11. **Find the values of \( r \)**:
\[ r = 9 \quad \text{or} \quad r = 4 \]

12. **Determine the valid solution**:
Since \( r \) must be less than or equal to 5 (as it is a permutation of 5), we discard \( r = 9 \). Therefore, the valid solution is:
\[ r = 4 \]

### Final Answer:
The value of \( r \) is \( 4 \).

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