Problem 41 - Algebra for JEE ADVANCED

Question

If the difference of the number of arrangements of three things from a certain number of dissimilar things and the number of selections of the same number of things from them exceeds 100 , then the least number of dissimilar things is a. 8 b. 6 c. 5 d. 7

Step-by-Step Solution

Answer

The least number of dissimilar things is 7 (option d).

Step 1: Identify the expressions for arrangements and combinations

The number of arrangements (permutations) of 3 things from a set of \( n \) dissimilar things is given by \( P(n, 3) = \frac{n!}{(n-3)!} \). The number of selections (combinations) is given by \( C(n, 3) = \frac{n!}{3!(n-3)!} \).

Step 2: Setup the inequality

We know from the problem that the difference between the number of arrangements and the number of selections exceeds 100. This means: \[P(n, 3) - C(n, 3) > 100\] Substitute the formulas: \[\frac{n!}{(n-3)!} - \frac{n!}{3!(n-3)!} > 100\]

Step 3: Simplify the inequality

Both terms have a common factor \( \frac{n!}{(n-3)!} \), which simplifies the inequality to: \[\frac{n!}{(n-3)!} \left( 1 - \frac{1}{6} \right) = \frac{5}{6} \cdot \frac{n!}{(n-3)!} > 100\] Multiply both sides by 6: \[5 \cdot \frac{n!}{(n-3)!} > 600\] Divide by 5: \[\frac{n!}{(n-3)!} > 120\]

Step 4: Calculate factorial values and test options for \( n \)

Now calculate the values: we need to find the smallest \( n \) such that \( \frac{n!}{(n-3)!} > 120 \). Try \( n=5 \):\[\frac{5!}{(5-3)!} = \frac{120}{2} = 60\] Try \( n=6 \):\[\frac{6!}{(6-3)!} = \frac{720}{6} = 120\] Try \( n=7 \):\[\frac{7!}{(7-3)!} = \frac{5040}{24} = 210\]Thus, for \( n=7 \), the condition is satisfied.

Key Concepts

Factorial

A factorial, denoted by an exclamation mark (!), is a mathematical operation that multiplies a series of descending natural numbers. To compute the factorial of a number, say \( n \), you multiply \( n \) by every positive whole number less than it down to 1. For example, if \( n = 5 \), then the factorial of 5, written as \( 5! \), is calculated as:\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]Factorials grow extremely fast. \( 4! \) is merely 24, but \( 6! \) already jumps up to 720. This rapid increase is why factorials are frequently used in probability and statistics, especially in permutations and combinations where the ordering or selection of items is involved. Factorials also appear frequently in equations, where simplifying them addresses expressions of descending order, like subtracting terms from a factorial divided by a factorial of a lesser number.

Inequalities

Inequalities represent a relationship where one number or expression is not equal to another, but instead is greater or less than another. In mathematical notation, inequalities are represented using symbols such as \(>\) (greater than), \(<\) (less than), \(\geq \) (greater than or equal to), and \(\leq \) (less than or equal to).
In our problem, the inequality \( P(n, 3) - C(n, 3) > 100 \) tells us that the difference between the permutations and combinations of 3 items from a group of \( n \) items must be greater than 100. Solving inequalities generally involves simplifying the expression to isolate the variable. With factorials involved, understanding simplification techniques is critical, like factoring common terms and simplifying fractions. This approach ensures that one can test different values confidently, ensuring the inequality holds true.

Arrangements and Selections

Arrangements refer to the ordering of items, which in mathematics is often addressed through permutations. When arranging items, every possible ordering is considered distinct. For example, if arranging the letters A, B, and C, both ABC and BAC are unique arrangements.
Selections, on the other hand, are combinations where the focus is on choosing items regardless of order. In combinations, ABC, BAC, and CAB are all treated as identical since the same three items are chosen.

• Permutations (arrangements) when selecting \( r \) items from \( n \) items, are calculated using \( P(n, r) = \frac{n!}{(n-r)!} \).
• Combinations (selections) are calculated as \( C(n, r) = \frac{n!}{r!(n-r)!} \).

Understanding this difference is essential in solving problems where both arrangements and selections are involved. In the exercise given, our task was to calculate both the permutations and combinations of 3 from \( n \), and understand how the arrangements exceeded the selections by more than 100. This highlights the importance of mathematical understanding in real-world scenarios, especially when determining the number of ways to choose or order items.