Solve for n :
` ((2n)!)/( 3!( 2n-3) !) : ( n!)/( 2!( n-2)!) = 44 : 3 `

Answer

To solve the equation

\[ \frac{(2n)!}{3!(2n-3)!} : \frac{n!}{2!(n-2)!} = \frac{44}{3}, \]

we will start by rewriting the ratio as an equation:

\[ \frac{(2n)!}{3!(2n-3)!} = \frac{44}{3} \cdot \frac{n!}{2!(n-2)!}. \]

### Step 1: Simplify the left-hand side

The left-hand side can be simplified using the definition of combinations:

\[ \frac{(2n)!}{3!(2n-3)!} = \binom{2n}{3}. \]

So we rewrite the equation as:

\[ \binom{2n}{3} = \frac{44}{3} \cdot \frac{n!}{2!(n-2)!}. \]

### Step 2: Simplify the right-hand side

The right-hand side can also be simplified:

\[ \frac{n!}{2!(n-2)!} = \binom{n}{2}. \]

Thus, we can rewrite the equation as:

\[ \binom{2n}{3} = \frac{44}{3} \cdot \binom{n}{2}. \]

### Step 3: Write out the combinations

Now we can express the combinations explicitly:

\[ \binom{2n}{3} = \frac{(2n)(2n-1)(2n-2)}{3!} = \frac{(2n)(2n-1)(2n-2)}{6}, \]

and

\[ \binom{n}{2} = \frac{n(n-1)}{2}. \]

### Step 4: Substitute back into the equation

Substituting these into our equation gives:

\[ \frac{(2n)(2n-1)(2n-2)}{6} = \frac{44}{3} \cdot \frac{n(n-1)}{2}. \]

### Step 5: Clear the fractions

To eliminate the fractions, multiply both sides by 6:

\[ (2n)(2n-1)(2n-2) = \frac{44 \cdot 6}{3} \cdot \frac{n(n-1)}{2}. \]

Calculating the right side:

\[ \frac{44 \cdot 6}{3} = 88. \]

So the equation simplifies to:

\[ (2n)(2n-1)(2n-2) = 88 \cdot \frac{n(n-1)}{2}. \]

### Step 6: Simplify further

This simplifies to:

\[ (2n)(2n-1)(2n-2) = 44n(n-1). \]

### Step 7: Expand both sides

Expanding the left-hand side:

\[ (2n)(2n-1)(2n-2) = 8n^3 - 12n^2 + 4n. \]

The right-hand side is:

\[ 44n^2 - 44n. \]

### Step 8: Set the equation to zero

Setting both sides equal gives:

\[ 8n^3 - 12n^2 + 4n - 44n^2 + 44n = 0. \]

Combining like terms:

\[ 8n^3 - 56n^2 + 48n = 0. \]

### Step 9: Factor out common terms

Factoring out \(n\):

\[ n(8n^2 - 56n + 48) = 0. \]

This gives us one solution \(n = 0\). Now we solve the quadratic:

\[ 8n^2 - 56n + 48 = 0. \]

### Step 10: Use the quadratic formula

Using the quadratic formula \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

\[ n = \frac{56 \pm \sqrt{(-56)^2 - 4 \cdot 8 \cdot 48}}{2 \cdot 8}. \]

Calculating the discriminant:

\[ 3136 - 1536 = 1600. \]

So,

\[ n = \frac{56 \pm 40}{16}. \]

Calculating the two possible values:

1. \(n = \frac{96}{16} = 6\)
2. \(n = \frac{16}{16} = 1\)

### Final Answer

Thus, the solutions for \(n\) are \(n = 6\) and \(n = 1\).

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