Question

The exponent of 7 in \( ^{100}C_{50} \) is :

• A. \( 4 \)
• B. \( 2 \)
• C. \( 1 \)
• D. \( 0 \) Correct Answer: (D) \( 0 \) Solution:

Hint:

Kummer's Theorem: The exponent of a prime \( p \) in \( ^nC_r \) is the number of "carries" when adding \( r \) and \( n-r \) in base \( p \).
For \( 50 + 50 \) in base 7:
\( 50 = (1 \cdot 7^2) + (0 \cdot 7^1) + (1 \cdot 7^0) = 101_7 \).
Adding \( 101_7 + 101_7 = 202_7 \).
Since there were 0 carries during the addition, the exponent is 0.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The "exponent of a prime \( p \) in \( n! \)" refers to the highest power of \( p \) that divides the factorial \( n! \). This is calculated using Legendre's Formula. Since binomial coefficients like \( ^nC_r \) are defined as fractions of factorials (\( \frac{n!}{r!(n-r)!} \)), the exponent of a prime in the coefficient is simply the exponent in the numerator minus the sum of the exponents in the denominator terms (based on the laws of exponents: \( p^a / (p^b \cdot p^c) = p^{a-b-c} \)).
This problem asks for the specific prime \( p = 7 \) in the context of \( ^{100}C_{50} \). This means we need to find how many times 7 appears as a factor in \( 100! \) and how many times it appears in \( (50!)^2 \).

Step 2: Key Formula or Approach:
Legendre's Formula: The exponent of prime \( p \) in \( n! \), denoted \( E_p(n!) \), is:
\[ E_p(n!) = \lfloor n/p \rfloor + \lfloor n/p^2 \rfloor + \lfloor n/p^3 \rfloor + \dots \]
where \( \lfloor x \rfloor \) is the floor function (greatest integer less than or equal to \( x \)).
For \( ^{n}C_{r} \), the exponent of \( p \) is:
\[ E_p(^nC_r) = E_p(n!) - E_p(r!) - E_p((n-r)!) \]

Step 3: Detailed Explanation:
Step 3.1: Calculate \( E_7(100!) \).
- \( \lfloor 100/7 \rfloor = 14 \) (There are 14 multiples of 7: 7, 14, ..., 98)
- \( \lfloor 100/49 \rfloor = 2 \) (There are 2 multiples of 49: 49, 98. Each contributes an extra factor of 7)
- \( \lfloor 100/343 \rfloor = 0 \)
Total \( E_7(100!) = 14 + 2 = 16 \).
Step 3.2: Calculate \( E_7(50!) \).
- \( \lfloor 50/7 \rfloor = 7 \) (Multiples: 7, 14, 21, 28, 35, 42, 49)
- \( \lfloor 50/49 \rfloor = 1 \) (Multiple: 49. It contributes an extra factor)
Total \( E_7(50!) = 7 + 1 = 8 \).
Step 3.3: Final Calculation for \( ^{100}C_{50} \).
The formula for the binomial coefficient is \( \frac{100!}{50! \cdot 50!} \).
The exponent of 7 is:
\[ E_7(^{100}C_{50}) = E_7(100!) - [E_7(50!) + E_7(50!)] \]
Substituting our calculated values:
\[ E_7(^{100}C_{50}) = 16 - [8 + 8] = 16 - 16 = 0 \]
This means that \( ^{100}C_{50} \) is not divisible by 7. In other words, 7 does not appear in its prime factorization.

Step 4: Final Answer:
The prime factor 7 appears exactly zero times in the expansion of \( ^{100}C_{50} \). This is because the number of factors of 7 in the denominator exactly cancels out those in the numerator. Thus, Option (D) is the correct answer.