The number of sequences of ten terms, whose terms are either 0 or 1 or 2, that contain exactly five

Q: Wait, the question never mentioned zeros! How can I assume there are two zeros?

That is the classic JEE trap! The problem states the sequence has ten terms. If you have five 1s and three 2s, that accounts for eight terms. The remaining $10 - 8 = 2$ terms must be 0s because the set of allowed digits is $\{0, 1, 2\}$. Always check the total count!

Q: Why do we divide by $5!$, $3!$, and $2!$? Why not just use $10!$?

Great question! $10!$ assumes all ten items are distinct. But here, the five 1s are indistinguishable. If you swap two 1s, the sequence looks exactly the same. Dividing by the factorial of the count of each identical group removes these redundant arrangements.

Q: Is there another way to solve this? Maybe using combinations?

Absolutely! You can think of it as choosing 5 slots for the 1s out of 10, then choosing 3 slots for the 2s out of the remaining 5, and the last 2 slots are automatically 0s. Mathematically, that is $\binom{10}{5} \times \binom{5}{3} \times \binom{2}{2}$, which simplifies to the same result!

Question

The number of sequences of ten terms, whose terms are either 0 or 1 or 2, that contain exactly five 1s and exactly three 2s, is equal to

Options:

(A) 360

(B) 45

(C) 2520

(D) 1820

Sigma Percentile · Accepted Answer

### The Art of Counting: Unlocking the Sequence

Welcome, future engineer. Today, we are not just solving a combinatorics problem; we are learning to see the hidden structure behind a chaotic string of numbers.

Imagine you are standing before a row of ten empty boxes. Your task is to fill them with the digits $0$, $1$, and $2$. But there is a catch—a strict set of rules that governs this arrangement.

Let us break this down together.

### Phase 1: The Hidden Deduction

The problem gives us two explicit conditions: we need exactly five $1$s and exactly three $2$s. If you stop here, you might feel like something is missing.

You have ten slots, but you have only accounted for eight of them ($5 + 3 = 8$). What about the remaining two slots?

Since the sequence can only contain $0$, $1$, or $2$, and we have already satisfied the requirements for $1$s and $2$s, the remaining two slots must be filled with $0$s.

This is the first step of a JEE master: always account for the 'invisible' constraints. We now have a set of ten objects: five $1$s, three $2$s, and two $0$s.

### Phase 2: The Multiset Permutation

Now, we face the core challenge. If all ten numbers were distinct, the number of arrangements would simply be $10!$.

But here, the items are identical within their groups. If you take a sequence and swap two $1$s, the sequence remains unchanged. This redundancy is the enemy of counting.

To fix this, we use the formula for permutations of multisets:

$$	ext{Number of sequences} = \frac{n!}{n_1! n_2! n_3!}$$

Here, $n = 10$ (the total slots), $n_1 = 5$ (the count of $1$s), $n_2 = 3$ (the count of $2$s), and $n_3 = 2$ (the count of $0$s). Substituting these values, we get:

$$	ext{Sequences} = \frac{10!}{5! 3! 2!}$$

### Phase 3: The Elegance of Cancellation

Do not let the factorials intimidate you. The beauty of mathematics lies in simplification.

Let us expand $10!$ just enough to cancel the largest factorial in the denominator, which is $5!$:

$$10! = 10 	imes 9 	imes 8 	imes 7 	imes 6 	imes 5!$$

Now, substitute this back into our expression:

$$	ext{Sequences} = \frac{10 	imes 9 	imes 8 	imes 7 	imes 6 	imes 5!}{5! 	imes 3! 	imes 2!}$$

The $5!$ terms vanish, leaving us with:

$$\frac{10 	imes 9 	imes 8 	imes 7 	imes 6}{3! 	imes 2!}$$

We know that $3! = 6$ and $2! = 2$. So the expression becomes:

$$\frac{10 	imes 9 	imes 8 	imes 7 	imes 6}{6 	imes 2}$$

Watch the magic happen: the $6$ in the numerator and denominator cancels out perfectly. Then, we divide $8$ by $2$ to get $4$.

We are left with a simple product:

$$10 	imes 9 	imes 4 	imes 7$$

### Phase 4: The Final Victory

Now, we perform the final arithmetic. $10 	imes 9 = 90$, and $4 	imes 7 = 28$.

Multiplying $90$ by $28$ gives us $2520$.

There it is—the total number of possible sequences is **$2520$**. You have successfully navigated the constraints, applied the correct combinatorial principle, and simplified the expression to reach the truth.

Keep this logic in your toolkit; the ability to identify identical objects in a set is a superpower in JEE Advanced combinatorics.

Animated Solution for Mathematics - Permutations and Combinations: The number of sequences of ten terms, whose terms are either 0 or 1 or 2, that contain exactly five 1s and exactly three 2s, is equal to

Select Answer:

Visualized Solution

Visualizing the $10$ slots

Placing five $1$ s

Placing three $2$ s

Deducing the number of $0$ s

Permutation of identical objects

The Multiset Formula

Expanding $10!$

Expanding $3!$ and $2!$

Final Multiplication

The Result

The Sigma Insight: Linear Permutations

The Art of Counting

Phase 1

Phase 2

Phase 3

Phase 4

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