In a shop there are five types of ice-creams available. A child buys six ice-creams. Statement-1 : T

Q: Wait, why is the formula $\binom{n+r-1}{r}$ and not just $\binom{n+r}{r}$?

Great question! Think of it this way: to divide $r$ items into $n$ categories, you need $n-1$ dividers (bars). So, you are arranging a total of $r$ stars and $n-1$ bars. The total number of positions is $r + (n-1) = n+r-1$. You need to choose $r$ positions for the stars (or $n-1$ for the bars). Hence, the formula is $\binom{n+r-1}{r}$.

Q: Why is Statement 1 false? Isn't $\binom{10}{5}$ the same as $\binom{10}{6}$?

Actually, no! While it is true that $\binom{n}{k} = \binom{n}{n-k}$, here we have $\binom{10}{6}$. The property $\binom{10}{6} = \binom{10}{10-6} = \binom{10}{4}$. Since $\binom{10}{4} \neq \binom{10}{5}$, Statement 1 is indeed false. Always double-check your symmetry properties!

Q: How does arranging 6 A's and 4 B's relate to the ice-cream problem?

Think of the 6 A's as your 6 ice-creams (stars) and the 4 B's as the 4 separators (bars) needed to create 5 categories. Arranging these 10 objects is exactly the same as choosing 6 positions for the A's out of 10 total spots, which is $\binom{10}{6}$. It's a beautiful way to visualize the same combinatorial logic!

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 $A$'s and 4 $B$'s in a row.

Options:

(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

Sigma Percentile · Accepted Answer

### The Ice-Cream Parlor Paradox

Imagine you walk into an ice-cream parlor. The air is sweet, the options are endless, and you have a mission: to buy exactly six ice-creams.

There are five delicious flavors available. Because you are buying six ice-creams and there are only five types, you are guaranteed to repeat at least one flavor.

This isn't just a snack run; it is a classic problem of selection with repetition. Let us dive into the mathematics of this choice.

### Translating Reality into Algebra

To solve this, we must first translate the physical situation into a rigorous mathematical equation. Let $x_1, x_2, x_3, x_4,$ and $x_5$ represent the number of ice-creams you buy of each type.

Since you are buying a total of six, we can write the equation:

$$x_1 + x_2 + x_3 + x_4 + x_5 = 6$$

Here, each $x_i$ must be a non-negative integer because you cannot buy a negative number of ice-creams. This is a classic linear Diophantine equation, and we need to find the number of non-negative integer solutions.

### The Stars and Bars Intuition

To solve this, we use the elegant 'Stars and Bars' method. Imagine your six ice-creams as six identical stars: $\star \star \star \star \star \star$.

We need to distribute these six stars into five distinct groups (the flavors). To create five groups, we need four separators, or 'bars': $|$.

For example, the arrangement $\star \star | \star | \star \star \star | |$ would mean you bought two of the first flavor, one of the second, three of the third, and zero of the fourth and fifth. To divide any set of items into $n$ groups, you always need $n-1$ bars. Here, $n=5$, so we need $5-1=4$ bars.

### The Combinatorial Masterpiece

Now, look at the total number of objects we are arranging. We have six stars and four bars, making a total of $6 + 4 = 10$ objects.

Every unique arrangement of these ten objects corresponds to a unique way of buying your ice-creams. The problem reduces to: in how many ways can we choose 6 positions for the stars out of 10 total positions?

This is simply the combination formula:

$$\binom{10}{6} = \frac{10!}{6!4!} = 210$$

Alternatively, if you chose the positions for the bars, you would get $\binom{10}{4}$, which is also 210. Both paths lead to the same beautiful truth.

### Evaluating the Statements

Now, let us look at the statements provided. Statement-1 claims the number of ways is $\binom{10}{5}$.

We just calculated that the correct answer is $\binom{10}{6}$. Since $\binom{10}{6} = 210$ and $\binom{10}{5} = 252$, **Statement-1 is clearly false**.

Statement-2 suggests that the number of ways is equal to the number of ways of arranging 6 A's and 4 B's. If we map our 6 stars to 6 A's and our 4 bars to 4 B's, we are indeed arranging 10 objects where 6 are identical and 4 are identical.

The number of permutations is:

$$\frac{10!}{6!4!} = \binom{10}{6}$$

Thus, **Statement-2 is true**.

### Final Reflection

Combinatorics is often about finding the right perspective. By visualizing the ice-creams as stars and the flavor-dividers as bars, we turned a daunting selection problem into a simple arrangement task.

Remember, in JEE Advanced, the trap is rarely in the calculation; it is in the conceptual leap. You have mastered the Stars and Bars method today—keep that intuition sharp, and no counting problem will ever stand in your way.

Select Answer:

Visualized Solution

Understanding the Problem

Formulating the Equation

The Stars and Bars Method

Adding the Bars

Total Objects to Arrange

Calculating Total Ways

Evaluating Statement-1

Evaluating Statement-2

Calculating Arrangements of A's and B's

Final Conclusion

The Sigma Insight: Combinations and Selection

The Ice-Cream Parlor Paradox

Translating Reality into Algebra

The Stars and Bars Intuition

The Combinatorial Masterpiece

Evaluating the Statements

Final Reflection

Similar Questions

Statement-1: The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box is empty is $^{9} C_{3}$ . Statement-2: The number of ways of choosing any 3 places from 9 different places is $^{9} C_{3}$ .

The number of ways to distribute 30 identical candies among four children $C_{1}, C_{2}, C_{3}$ and $C_{4}$ so that $C_{2}$ receives atleast 4 and atmost 7 candies, $C_{3}$ receives atleast 2 and atmost 6 candies, is equal to

A boy needs to select five courses from 12 available courses, out of which 5 courses are language courses. If he can choose at most two language courses, then the number of ways he can choose five courses is ____.

The number of ways, in which 16 oranges can be distributed to four children such that each child gets at least one orange, is

Six X's have to be placed in the squares of figure below in such a way that each row contains at least one X. In how many different ways can this be done.

Number of ways of arranging 8 identical books into 4 identical shelves where any number of shelves may remain empty is equal to

The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets at least one ball is

If $^{n} C_{r}$ denotes the number of combination of $n$ things taken $r$ at a time, then the expression $^{n} C_{r + 1} +^{n} C_{r - 1} + 2 \times^{n} C_{r}$ equals