Step 1: Understanding the Concept:
We use the symmetric property of binomial coefficients which states \(\binom{n}{k} = \binom{n}{n-k}\).
This property allows us to simplify the given expression into a much more recognizable form.
Step 2: Key Formula or Approach:
Apply the symmetry property to each term individually.
\(\binom{30}{30-r} = \binom{30}{r}\)
\(\binom{30}{31-r} = \binom{30}{r-1}\)
\(\binom{30}{32-r} = \binom{30}{r-2}\)
\(\binom{30}{33-r} = \binom{30}{r-3}\)
The given expression simplifies to the following sum.
\[ S = \binom{30}{r} + 3\binom{30}{r-1} + 3\binom{30}{r-2} + \binom{30}{r-3} \]
Step 3: Detailed Explanation:
Notice that the coefficients \(1, 3, 3, 1\) are exactly the binomial coefficients of \((1+x)^3\).
This suggests finding the coefficient of \(x^r\) in a product of two polynomials.
Consider the algebraic identity:
\[ (1+x)^{30} \cdot (1+x)^3 = (1+x)^{33} \]
Let's find the coefficient of \(x^r\) on both sides of this identity.
On the left side, we expand both binomials.
\[ \left( \dots + \binom{30}{r}x^r + \binom{30}{r-1}x^{r-1} + \dots \right) \times \left( \binom{3}{0} + \binom{3}{1}x + \binom{3}{2}x^2 + \binom{3}{3}x^3 \right) \]
The term with \(x^r\) is obtained by multiplying pairs that give a combined degree of \(r\).
\[ \binom{30}{r}\binom{3}{0} + \binom{30}{r-1}\binom{3}{1} + \binom{30}{r-2}\binom{3}{2} + \binom{30}{r-3}\binom{3}{3} \]
Substitute the known values of \(\binom{3}{k}\).
\[ = \binom{30}{r}(1) + \binom{30}{r-1}(3) + \binom{30}{r-2}(3) + \binom{30}{r-3}(1) \]
This matches our simplified expression \(S\) perfectly.
On the right side of the identity \((1+x)^{33}\), the coefficient of \(x^r\) is simply \(\binom{33}{r}\).
Therefore, we can equate the two representations.
\[ \binom{33}{r} = \binom{m}{r} \implies m = 33 \]
Step 4: Final Answer:
The value of \(m\) is \(33\).