Key Concepts and Formulas

• Combinations: The number of ways to choose $r$ items from a set of $n$ distinct items without regard to order, denoted by $\binom{n}{r}$ or $^nC_r$, and calculated as $\binom{n}{r} = \frac{n!}{r!(n-r)!}$.
• Strict Inequality and Distinctness: The condition $n_1 < n_2 < n_3$ implies that $n_1, n_2,$ and $n_3$ must be distinct.
• Factorial: The product of all positive integers less than or equal to a given positive integer $n$, denoted by $n!$, and calculated as $n! = n \times (n-1) \times (n-2) \times \dots \times 2 \times 1$.

Step-by-Step Solution

Step 1: Understand the Problem and the Constraint We are asked to find the number of ways to select three numbers $n_1, n_2, n_3$ from the set $\{1, 2, \dots, 10\}$ such that $n_1 < n_2 < n_3$. The strict inequality forces the three numbers to be distinct, and it also imposes a unique order on them.

Step 2: Reframe the Problem as a Combination Because the order of the three numbers is already determined by the strict inequality, we only need to find the number of ways to choose three distinct numbers from the set of ten. This is a combination problem.

Step 3: Apply the Combination Formula We have $n = 10$ (the total number of distinct items) and $r = 3$ (the number of items to choose). Therefore, the number of ways to choose the numbers is given by $\binom{10}{3}$.

Step 4: Calculate the Combination Using the combination formula, we have: $\binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10!}{3!7!}$

Step 5: Simplify the Factorials Expand the factorials and simplify: $\binom{10}{3} = \frac{10 \times 9 \times 8 \times 7!}{3 \times 2 \times 1 \times 7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1}$

Step 6: Calculate the Result Perform the multiplication and division: $\binom{10}{3} = \frac{720}{6} = 120$

Thus, there are 120 ways to choose the balls such that $n_1 < n_2 < n_3$ if the range was 1 to 10. However, we are given that the answer is 164. This implies an error in the question or the given answer. Let us assume that the balls are labeled 0 to 10 and we want to select the balls such that $n_1 < n_2 < n_3$. Then we would have 11 choices and we would be choosing 3 balls. This would lead to:

Step 7: Assume Balls are labeled 0 to 10 and we want to select the balls such that $n_1 < n_2 < n_3$ We have $n = 11$ (the total number of distinct items) and $r = 3$ (the number of items to choose). Therefore, the number of ways to choose the numbers is given by $\binom{11}{3}$. Using the combination formula, we have: $\binom{11}{3} = \frac{11!}{3!(11-3)!} = \frac{11!}{3!8!}$ Expanding the factorials and simplify: $\binom{11}{3} = \frac{11 \times 10 \times 9 \times 8!}{3 \times 2 \times 1 \times 8!} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1}$ Performing the multiplication and division: $\binom{11}{3} = \frac{990}{6} = 165$

The result is 165, which is close to the answer of 164. Let's assume that one of the number can also be equal. Then, if $n_1 \le n_2 < n_3$, we can pick the numbers in $\binom{10}{2}$ ways, where two numbers are equal, but this does not fit the scenario where $n_1, n_2, n_3$ are balls from three different boxes. Let us assume that the first box has 9 balls, the second box has 10 balls, and the third box has 11 balls.

Then we have to choose $n_1$ from 1 to 9, $n_2$ from 1 to 10, and $n_3$ from 1 to 11. If $n_1=1$, then $n_2$ can be anything from 2 to 10. If $n_2=2$, $n_3$ can be 3 to 11. If $n_2=10$, $n_3$ can be 11. If $n_1=9$, then $n_2$ can be 10, and $n_3$ has to be 11. $\sum_{n_1=1}^{9} \sum_{n_2=n_1+1}^{10} \sum_{n_3=n_2+1}^{11} 1$

$\sum_{n_1=1}^{9} \sum_{n_2=n_1+1}^{10} 11 - n_2 = \sum_{n_1=1}^{9} \sum_{n_2=n_1+1}^{10} 11 - (n_1+1) - (10-n_2)$ $\sum_{n_1=1}^{9} \sum_{k=1}^{10-n_1} (11-(n_1+k)) = \sum_{n_1=1}^{9} \sum_{k=1}^{10-n_1} 11 - n_1 - k$ $\sum_{n_1=1}^{9} (11-n_1)(10-n_1) - \frac{(10-n_1)(11-n_1)}{2}$

The correct answer provided must be incorrect. $\binom{10}{3} = 120$.

Common Mistakes & Tips

• Confusing Combinations and Permutations: Remember that combinations are used when the order doesn't matter, while permutations are used when the order does matter. In this case, the strict inequality dictates the order, so we use combinations.
• Incorrectly Applying the Formula: Double-check that you have correctly identified $n$ (the total number of items) and $r$ (the number of items to choose).
• Forgetting the Distinctness Condition: The strict inequality $n_1 < n_2 < n_3$ implies that the numbers must be distinct.

Summary

The problem asks for the number of ways to select three distinct numbers from the set $\{1, 2, \dots, 10\}$ such that they are in strictly increasing order. Because the order is determined by the strict inequality, the problem reduces to finding the number of combinations of choosing 3 items from 10, which is $\binom{10}{3} = 120$. The provided answer of 164 is incorrect.

Final Answer The final answer is \boxed{120}, which corresponds to option (D).