Let P be the set of seven digit numbers with sum of their digits equal to 11. If the numbers in P ar

Q: Wait, why can't we just use stars and bars here? It looks like a distribution problem.

Stars and bars is excellent for distributing identical items into distinct bins, but here, the 'bins' (the 7 positions in the number) are distinct, and the 'items' (the digits) have specific values that contribute to a sum. Because the digits have different weights (1, 2, and 3), the sum constraint forces us to find the specific counts of each digit first, rather than just distributing them blindly.

Q: How do I know I haven't missed a case for (x, y, z)?

Great question! By reducing the system to y + 2z = 4, we constrained the problem significantly. Since y and z must be non-negative integers, z can only be 0, 1, or 2. If z were 3, y would be -2, which is impossible. This systematic approach ensures that every possible combination is captured without overlap.

Q: Why do we divide by factorials like 5! and 2! in the permutation step?

That is the beauty of permutations with repetition! When you have identical digits (like five 1s), swapping two 1s doesn't create a new number. Dividing by the factorial of the count of each repeated digit 'divides out' those redundant arrangements, leaving you with only the unique, distinct numbers.

Q: Is there a way to solve this using generating functions?

Absolutely! You could look for the coefficient of x^11 in the expansion of (x^1 + x^2 + x^3)^7. While that is a powerful, elegant method, it is often more time-consuming for a 7-digit problem than the case-based approach. For JEE, always choose the path that minimizes calculation errors!

Q: What if the sum was much larger? Would this method still work?

If the sum were larger, the number of cases would increase, and the case-based method might become tedious. In such scenarios, the generating function approach or a recurrence relation would be more robust. However, for small constraints like these, the case-based method is the safest and most intuitive way to ensure accuracy.

Question

Let P be the set of seven digit numbers with sum of their digits equal to 11. If the numbers in P are formed by using the digits 1, 2 and 3 only, then the number of elements in the set P is:

Options:

(A) 173

(B) 164

(C) 158

(D) 161

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We define our variables as $x$, $y$, and $z$, representing the frequency of the digits 1, 2, and 3, respectively. We are constrained by two fundamental equations.

First, the total number of digits is 7:

$$x + y + z = 7$$

Second, the sum of these digits must be 11:

$$1x + 2y + 3z = 11$$

By subtracting the first equation from the second, we derive the master key for our system:

$$y + 2z = 4$$

Since $x$, $y$, and $z$ must be non-negative integers, we can now systematically explore the possible scenarios for these variables.

### The Three Possible Worlds

We test the possible values of $z$ (the number of 3s) to determine the corresponding values for $y$ (the number of 2s) and $x$ (the number of 1s).

**Case 1: The World of 1s and 3s ($z=2$)**

If $z = 2$, then $y + 2(2) = 4$, which implies $y = 0$. Substituting these into our total count, we find $x + 0 + 2 = 7$, so $x = 5$.

The multiset is $\{1, 1, 1, 1, 1, 3, 3\}$. The number of distinct arrangements is given by the multinomial coefficient:

$$	ext{Ways} = \frac{7!}{5!2!} = \frac{7 	imes 6}{2} = 21$$

**Case 2: The World of 1s, 2s, and 3s ($z=1$)**

If $z = 1$, then $y + 2(1) = 4$, which implies $y = 2$. Substituting these into our total count, we find $x + 2 + 1 = 7$, so $x = 4$.

The multiset is $\{1, 1, 1, 1, 2, 2, 3\}$. The number of distinct arrangements is:

$$	ext{Ways} = \frac{7!}{4!2!1!} = \frac{7 	imes 6 	imes 5}{2} = 105$$

**Case 3: The World of 1s and 2s ($z=0$)**

If $z = 0$, then $y = 4$. Substituting these into our total count, we find $x + 4 + 0 = 7$, so $x = 3$.

The multiset is $\{1, 1, 1, 2, 2, 2, 2\}$. The number of distinct arrangements is:

$$	ext{Ways} = \frac{7!}{3!4!} = \frac{7 	imes 6 	imes 5}{3 	imes 2 	imes 1} = 35$$

### Final Calculation

Because these cases are mutually exclusive, we sum the results from each scenario to find the total number of valid seven-digit numbers.

$$	ext{Total} = 21 + 105 + 35 = 161$$

The total number of such seven-digit numbers is **161**.

Animated Solution for Mathematics - Permutations and Combinations: Let P be the set of seven digit numbers with sum of their digits equal to 11. If the numbers in P are formed by using the digits 1, 2 and 3 only, then the number of elements in the set P is:

Select Answer:

Visualized Solution

Problem Analysis & Constraints

Formulating the Mathematical Model

Case 1: Using Digits $1$ and $3$

Permutations for Case 1

Case 2: Using Digits $1, 2,$ and $3$

Permutations for Case 2

Case 3: Using Digits $1$ and $2$

Permutations for Case 3

Total Number of Elements in $P$

The Sigma Insight: Linear Permutations

Analyzing the Setup

The Three Possible Worlds

Final Calculation

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