If `.^(15)C_(r): .^(15)C_(r-1) = 1:5`, then find r.

Answer

To solve the problem, we start with the equation given:

\[ \frac{^{15}C_r}{^{15}C_{r-1}} = \frac{1}{5} \]

**Step 1: Use the formula for combinations.**

Recall the formula for combinations:

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

Substituting \( n = 15 \):

\[ ^{15}C_r = \frac{15!}{r!(15-r)!} \]
\[ ^{15}C_{r-1} = \frac{15!}{(r-1)!(15-(r-1))!} = \frac{15!}{(r-1)!(16-r)!} \]

**Step 2: Substitute the combinations into the equation.**

Now substitute these into the equation:

\[ \frac{\frac{15!}{r!(15-r)!}}{\frac{15!}{(r-1)!(16-r)!}} = \frac{1}{5} \]

**Step 3: Simplify the equation.**

The \( 15! \) cancels out:

\[ \frac{(16-r)!}{(15-r)!} \cdot \frac{(r-1)!}{r!} = \frac{1}{5} \]

Now, simplify \( \frac{(16-r)!}{(15-r)!} = 16 - r \) and \( \frac{(r-1)!}{r!} = \frac{1}{r} \):

\[ (16 - r) \cdot \frac{1}{r} = \frac{1}{5} \]

**Step 4: Cross-multiply to eliminate the fraction.**

Cross-multiplying gives:

\[ 5(16 - r) = r \]

**Step 5: Expand and rearrange the equation.**

Expanding this gives:

\[ 80 - 5r = r \]

Now, rearranging the terms:

\[ 80 = r + 5r \]
\[ 80 = 6r \]

**Step 6: Solve for \( r \).**

Now, divide both sides by 6:

\[ r = \frac{80}{6} = \frac{40}{3} \]

Thus, the value of \( r \) is:

\[ r = \frac{40}{3} \]

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