Problem 118 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If ' \(n\) 'is an integer between 0 and 21 , then the minimum value of \(n !(21-n) !\) is (A) \(9 ! 2 !\) (B) \(10 ! 11 !\) (C) \(20 !\) (D) \(21 !\)

Step-by-Step Solution

Answer

Minimum value is \(10!11!\).

Step 1: Understand the Problem

We need to find the minimum value of the expression \( n!(21-n)! \) for integer values of \( n \) ranging between 0 and 21. Each distinct \( n \) will give a different factorization, which we need to evaluate.

Step 2: Identify Symmetry

Observe that the expression \( n!(21-n)! \) exhibits symmetry around \( n = 10.5 \) because \( n = k \) and \( n = 21-k \) will yield the same value. Thus, analyzing these values will help find the minimum value.

Step 3: Evaluate the Middle Values

As identified, evaluate \( n = 10 \) and \( n = 11 \):- For \( n = 10 \): \( 10!(21-10)! = 10!11! \).- For \( n = 11 \): \( 11!(21-11)! = 11!10! \), which simplifies the same as \( 10!11! \).Both are evaluated to see which gives a smaller factorial product.

Step 4: Check Boundary Values

Compare values at the extremes, such as \( n = 0, 21 \). Both give:- \( n = 0 \): \( 0!21! = 21! \)- \( n = 21 \): \( 21!0! = 21! \)These are larger than \( 10!11! \).

Step 5: Compare Results

From Steps 3 and 4, \( 10!(21-10)! = 10!11! \) was found to be the smallest value when compared with boundary and other values. Thus, \( n = 10 \) or \( 11 \) yields the minimum value.

Key Concepts

Factorial Symmetry

Exploring the symmetry in factorial expressions can simplify complex arithmetic evaluations. In this exercise, we focus on the expression \( n!(21-n)! \). A notable characteristic of this expression is its symmetry around the center point, which is \( n = 10.5 \). This symmetry means that the factorials on either side of the center have an equal product.
To understand this, consider any integer \( n=k \) and its complement \( n=21-k \). The expressions \( k!(21-k)! \) and \( (21-k)!k! \) are equivalent due to the commutative property of multiplication, making the product symmetric around the midpoint of the range.
This symmetry is extremely useful when computing factorials. It means that examining values around \( n=10 \) and \( n=11 \) is sufficient to find the minimum value in the given range. This reduces the need for excessive computation.

Minimum Value

Finding the minimum value of the expression \( n!(21-n)! \) is the heart of solving the given exercise. Given the factorial symmetry around \( n=10.5 \), identifying the smallest product involves mainly analyzing values near the middle of the range.
Two central values are \( n=10 \) and \( n=11 \):

• For \( n=10 \), we compute \( 10!(21-10)! = 10!11! \).
• For \( n=11 \), the computation is \( 11!(21-11)! = 11!10! \), which is identical to the previous result.

Both evaluations result in the same product, emphasizing the minimum values lie around \( n=10 \) or \( n=11 \).
On comparing with boundary values such as \( n=0 \) and \( n=21 \), resulting in \( 0!21! = 21! \) and \( 21!0! = 21! \) respectively, it is clear that \( 10!11! \) yields the smallest possible factorial product.

Integer Evaluation

Evaluating the expression \( n!(21-n)! \) for integer values requires a strategic approach to select values efficiently and reduce computational effort. By focusing on critical integer points around the symmetry, calculations become more manageable.
The key integers here are \( n=10 \) and \( n=11 \), which appear as optimal values due to their central placement in the range. For each integer \( n \) evaluated, different expressions result due to changing factorial components, yet symmetry ensures efficiency.
Additionally, evaluating the integer boundary points \( n=0 \) and \( n=21 \) helps confirm these are not optimal, as both result in larger factorial products \( 21! \).
Thus, integer evaluation of this factorial problem boils down to careful analysis of mid-range points, making it clearer why \( n=10 \) or \( n=11 \) offer the minimum factorial value.