Problem 53 - A Complete Resource Book in Mathematics for JEE Main

Question

In a shop there are five types of ice-creams available. A child buys six ice- creams. Statement 1: The number of different ways the child can buy the six ice-creams is \({ }^{10} C_{5}\). Statement 2: The number of different ways the child can buy the six ice-creams is equal to the number of different ways of arranging \(6 \mathrm{~A}\) 's and \(4 \mathrm{~B}\) 's in a row. (A) Statement 1 is false, Statement 2 is true (B) Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1 (C) Statement 1 is true, Statement 2 is true; Statement 2 is not a correct explanation for Statement 1 (D) Statement 1 is true, Statement 2 is false

Step-by-Step Solution

Answer

(B) Statement 1 is true, Statement 2 is true, and Statement 2 correctly explains Statement 1.

Step 1: Interpret the Problem

We are given a situation where a child wants to buy 6 ice-creams from 5 different types available in a shop. We need to determine if the ways to do so can be expressed with the given mathematical statements.

Step 2: Analyze Statement 1

Statement 1 claims that the number of ways to purchase these 6 ice-creams is represented by the combination \(^{10}C_{5}\). In the context of combinatorics, we use the formula for combinations with repetition: \[ \ \binom{n+r-1}{r} = \binom{5+6-1}{6} = \binom{10}{6} = \binom{10}{4} \\] Which evaluates to \(\binom{10}{4}\). Therefore, Statement 1 is true.

Step 3: Analyze Statement 2

Statement 2 equates the number of ways to buy 6 ice-creams to arranging 6 A's and 4 B's in a row, implying 10 total items with 6 being one type. For arrangements, formula used is \[ \frac{10!}{6!\times 4!} = \binom{10}{4} \] This matches the result from the combination formula used in Statement 1, confirming Statement 2 is true.

Step 4: Determine Correct Explanation

Statement 2 correctly suggests a combinatorial equivalence that arranging 6 A's and 4 B's represents selecting different types and could serve as an explanation of finding combinations with repetition methodology applied in Statement 1.

Step 5: Conclude with the Correct Option

Since both statements are true and Statement 2 provides a correct reasoning for Statement 1 as it directly relates to the "stars and bars" theorem approach, the correct answer is option (B).

Key Concepts

Combinations with Repetition

When deciding on how many ways you can choose certain items from a set, combinations with repetition might come in handy. Imagine you want to select scoops of ice-cream where the order doesn’t matter, and you are allowed to choose the same flavor more than once. That's where combinations with repetition become relevant.

Mathematically, to calculate how many ways you can choose \( r \) items from \( n \) different items, where each can be selected more than once, we use the formula:

• \[ \binom{n+r-1}{r} \]

This formula helps us account for repeated choices and can be particularly useful in situations like buying ice-creams from limited options. In our ice-cream shop example, with 5 types of ice-creams and choosing 6, the formula becomes \( \binom{10}{4} \). The result shows us the number of combinations when order doesn’t matter, but repetitions are allowed.

Binomial Coefficient

The binomial coefficient is a key concept in combinatorics, represented by \( \binom{n}{k} \), and it signifies the number of ways to choose \( k \) objects from a set of \( n \) objects, without regard to order. It is an essential tool in situations where you're determining ways to combine items.

The formula to calculate binomial coefficients is:

• \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]

In our exercise, the process of arranging 6 A's and 4 B's, which follows binomial coefficient calculation, describes the selection methods among various choices, ensuring that both arrangements and repetitions are properly accounted for. Hence, translating the total selection of 10 into respective parts of 6 and 4 is simplified through the understanding of binomial coefficients.

Stars and Bars Theorem

The "stars and bars" theorem is a fundamental principle often used to solve problems of placing indistinguishable items into distinguishable boxes. In simpler terms, it helps determine how many ways \( n \) identical items can be distributed into \( k \) distinct groups or types.

The idea can be understood as envisioning items as "stars" and dividers as "bars". For example, wanting to distribute 6 ice-creams among 5 types, we conceptualize the ice-creams as stars and determine the number of ways we can place 4 dividers (bars) among them to separate into 5 groups. The general formula here aligns with combinations with repetition:

• \[ \binom{n+k-1}{k-1} \]

In the exercise context, arranging 6 A's and 4 B's illustrates the use of this theorem, bridging the gap between distribution and combination with repetition, demonstrating neatly how an abstract combinatorial theorem can be incorporated into practical calculations.