Let S denote the set of 4-digit numbers abcd such that a b c d and P denote the set of 5-digit

Q: Wait, why do we use $\binom{10}{4}$ for set $S$? Shouldn't we consider permutations?

That is a fantastic question! The condition $a > b > c > d$ is the key. Once you select any 4 distinct digits, there is only one possible way to arrange them in strictly decreasing order. For example, if you pick $\{1, 3, 7, 9\}$, the only valid number is $9731$. Therefore, the number of arrangements is fixed at $1$ for every combination. We only need to count the number of ways to choose the digits, which is $\binom{10}{4}$.

Q: Why can't we include $0$ in the digits for set $P$?

Excellent observation. The problem states the product of the 5 digits must be $20$. If any digit were $0$, the product of all 5 digits would be $0$, not $20$. Thus, $0$ is excluded from our set of possible digits.

Q: How do I know I've found all possible combinations for set $P$?

We rely on the prime factorization of $20$, which is $2^2 \times 5$. To get a product of $20$ using 5 digits, we must distribute these factors. We have one $5$ and two $2$s. We can either keep the two $2$s separate (as $2$ and $2$) or combine them into a $4$. Any remaining slots must be filled with $1$s to maintain the product. This systematic approach ensures we cover all possibilities: $\{5, 4, 1, 1, 1\}$ and $\{5, 2, 2, 1, 1\}$.

Question

Let S denote the set of 4-digit numbers $abcd$ such that $a > b > c > d$ and P denote the set of 5-digit numbers having product of its digits equal to 20. Then $n(S) + n(P)$ is equal to ......... .

Sigma Percentile · Accepted Answer

### Analyzing the Mystery of Set $S$

Imagine you are standing before a row of four empty slots, waiting to be filled by digits $a, b, c,$ and $d$. The condition is strict: $a > b > c > d$.

This condition is a gift; it implies that the order of any four chosen digits is already fixed. If we pick any four distinct digits from the set $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$, there is exactly one way to arrange them to satisfy the inequality.

The problem reduces to a simple selection of 4 distinct digits from the 10 available. The number of ways to do this is given by the combination formula:

$$n(S) = \binom{10}{4} = \frac{10 	imes 9 	imes 8 	imes 7}{4 	imes 3 	imes 2 	imes 1} = 210$$

### Analyzing the Mystery of Set $P$

Now, we consider set $P$, which consists of five-digit numbers where the product of the digits is exactly $20$. Since the product is non-zero, none of the digits can be $0$.

The prime factorization of $20$ is $2^2 	imes 5$. To fill five slots such that their product is $20$, we must use the factors $5, 2, 2$ and fill the remaining two slots with $1$s.

We identify two distinct cases based on the distribution of these factors:

**Case 1:** We combine the two $2$s into a $4$. The set of digits is $\{5, 4, 1, 1, 1\}$. The number of permutations is:

$$\frac{5!}{3!} = 20$$

**Case 2:** We keep the two $2$s separate. The set of digits is $\{5, 2, 2, 1, 1\}$. The number of permutations is:

$$\frac{5!}{2!2!} = 30$$

Summing these cases, we find the total number of elements in set $P$:

$$n(P) = 20 + 30 = 50$$

### The Final Synthesis

We have successfully determined the cardinality of both sets. We found $n(S) = 210$ and $n(P) = 50$.

The final step is the summation of these two values:

$$n(S) + n(P) = 210 + 50 = 260$$

The final result is **260**. By breaking down the constraints, we transformed a complex counting problem into a series of logical, manageable steps.

Animated Solution for Mathematics - Permutations and Combinations: Let S denote the set of 4-digit numbers $ab c d$ such that $a > b > c > d$ and P denote the set of 5-digit numbers having product of its digits equal to 20. Then $n (S) + n (P)$ is equal to ......... .

Enter Numerical Value:

Visualized Solution

Introduction to Set $S$

The Decreasing Order Condition

Calculating $n (S)$

Introduction to Set $P$

Factorizing the Product

Case 1: Digits ${5, 4, 1, 1, 1}$

Case 2: Digits ${5, 2, 2, 1, 1}$

Total Elements in Set $P$

Final Sum

The Sigma Insight: Combinations and Selection

Analyzing the Mystery of Set $S$

Analyzing the Mystery of Set $P$

The Final Synthesis

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Enter Numerical Value:

Visualized Solution

Introduction to Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

The Decreasing Order Condition

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n(S)

Introduction to Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P