If `""^(18)C_(15)+2(""^(18)C_(16))+""^(17)C_(16)+1=""^(n)C_(3)`, then n is equal to :

Options

A. 19
B. 20
C. 18
D. 24

Answer

To solve the equation \( \binom{18}{15} + 2\binom{18}{16} + \binom{17}{16} + 1 = \binom{n}{3} \), we will simplify the left-hand side step by step.

### Step 1: Rewrite the binomial coefficients
We can rewrite the binomial coefficients using the property \( \binom{n}{r} = \binom{n}{n-r} \):
\[ \binom{18}{15} = \binom{18}{3}, \quad \binom{18}{16} = \binom{18}{2}, \quad \text{and} \quad \binom{17}{16} = \binom{17}{1} \]
Thus, we can rewrite the equation as:
\[ \binom{18}{3} + 2\binom{18}{2} + \binom{17}{1} + 1 \]

### Step 2: Substitute the values
Now, we substitute the values:
\[ \binom{17}{1} = 17 \]
So our equation becomes:
\[ \binom{18}{3} + 2\binom{18}{2} + 17 + 1 \]
This simplifies to:
\[ \binom{18}{3} + 2\binom{18}{2} + 18 \]

### Step 3: Use the identity for binomial coefficients
We can use the identity:
\[ \binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1} \]
to combine the terms. First, we rewrite \( 2\binom{18}{2} \) as:
\[ \binom{18}{2} + \binom{18}{2} \]
Now we can combine:
\[ \binom{18}{3} + \binom{18}{2} + \binom{18}{2} + 18 = \binom{18}{3} + \binom{18}{2} + \binom{18}{2} + \binom{17}{1} \]
Using the identity:
\[ \binom{18}{3} + 2\binom{18}{2} + 17 + 1 = \binom{19}{3} \]

### Step 4: Set the equation
Now we have:
\[ \binom{19}{3} = \binom{n}{3} \]

### Step 5: Solve for \( n \)
From the equality of binomial coefficients, we can conclude:
\[ n = 19 \]

Thus, the value of \( n \) is \( 20 \).

### Final Answer:
The value of \( n \) is \( 20 \).

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