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 ________ .
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Solution Steps
1
Calculating $n(S)$ for Set S
Set $S$ consists of 4-digit numbers $abcd$ with $a > b > c > d$. The digits are chosen from $\{0, 1, 2, …, 9\}$.
Since the digits must be strictly decreasing, any choice of 4 distinct digits will result in exactly 1 valid number.
$n(S) = \binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$.
2
Case Analysis for Set P
Set $P$ consists of 5-digit numbers whose product of digits is 20. Prime factorization of 20 is $2^2 \times 5$.
The possible combinations of 5 digits (excluding 0) that multiply to 20 are:
- Case 1: $\{5, 4, 1, 1, 1\}$
- Case 2: $\{5, 2, 2, 1, 1\}$
3
Permutations for Case 1
For digits $\{5, 4, 1, 1, 1\}$, the number of arrangements is:
$\frac{5!}{3!} = \frac{120}{6} = 20$.
4
Permutations for Case 2
For digits $\{5, 2, 2, 1, 1\}$, the number of arrangements is:
$\frac{5!}{2! \times 2!} = \frac{120}{4} = 30$.
5
Total for Set P and Final Sum
$n(P) = 20 + 30 = 50$.
The required value is $n(S) + n(P) = 210 + 50 = 260$.
Final Result: 260
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Theory
1. Selection for Monotonic Sequences
When we need to form numbers where digits follow a strictly increasing or decreasing order, the process is simplified to a combination problem. For a sequence of $r$ digits chosen from $n$ available digits such that they are strictly decreasing ($d_1 > d_2 > … > d_r$), any set of $r$ distinct digits selected can be arranged in exactly one way to satisfy the condition. Therefore, the total number of such numbers is simply $\binom{n}{r}$. Note that if the condition were strictly increasing, the digit 0 could not be selected as the leading digit.
2. Product of Digits and Prime Factorization
To find sets of digits that produce a specific product, we rely on the prime factorization of that product. For a product of 20 ($2^2 \times 5$), the digit 5 must be present because it is a prime factor greater than 3. The remaining factors ($2^2 = 4$) can be grouped as a single digit (4) or two digits (2 and 2). Since we need a 5-digit number, the remaining places must be filled with the digit 1, as 1 is the multiplicative identity. This systematic approach ensures no combinations are missed.
3. Permutations of Multisets
When calculating the number of ways to arrange digits where some digits are identical, we use the formula for permutations of a multiset: $n! / (p_1! p_2! … p_k!)$, where $n$ is the total number of items and $p_i$ is the frequency of each identical item. In Case 1 of set P, we have three 1s, leading to a divisor of $3!$. In Case 2, we have two 2s and two 1s, leading to a divisor of $2! \times 2!$. This is a fundamental concept in counting distinct arrangements.
4. Constraint Satisfaction in Number Formation
In combinatorial problems involving numbers, “constraints” define the search space. In set S, the constraint is a relative order ($a > b > c > d$). In set P, the constraint is an algebraic product ($d_1 \cdot d_2 \cdot d_3 \cdot d_4 \cdot d_5 = 20$). When solving JEE problems, identifying these constraints and translating them into mathematical operations (like $\binom{n}{r}$ or multinomial coefficients) is the primary goal. Always check if the leading digit can be 0; in this problem, the product condition (20) and the decreasing order ($a > b > c > d$) automatically prevent the leading digit from being 0.
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FAQs
1
Why did we not consider the case with digit 10 or 20?
Digits in a number can only range from 0 to 9. Therefore, factors of the product must be single-digit integers.
2
Why does 10C4 include 0?
For strictly decreasing digits $a>b>c>d$, if 0 is selected, it must be $d$. Since $a>b>c>d \ge 0$, the leading digit $a$ will always be at least 3, so it remains a 4-digit number.
3
Could there be a case for P like {10, 2, 1, 1, 1}?
No, because 10 is not a single digit. Every “digit” in a 5-digit number must be $\le 9$.
4
What if the condition was a ≥ b ≥ c ≥ d?
In that case, we would use the formula for combinations with repetition, $\binom{n+r-1}{r}$.
5
Is the number 55111 possible for P?
No, the product $5 \times 5 \times 1 \times 1 \times 1 = 25$, which is not 20.
6
How many 5-digit numbers have a product of digits equal to 0?
This would be (Total 5-digit numbers) – (Numbers with no 0 digit) = $90,000 – 9^5$.
7
What is the formula for nCr?
$n! / (r!(n-r)!)$. For 10C4, it is $10! / (4! \times 6!)$.
8
Why are there only two cases for set P?
Factorizing 20 into 5 integers $\le 9$: the prime factor 5 must be one digit. The remaining $2 \times 2$ can only be $\{4,1,1,1\}$ or $\{2,2,1,1\}$.
9
Does n(P) include numbers starting with 1?
Yes, numbers like 11145 or 12125 are valid 5-digit numbers and belong to set P.
10
What if the question asked for n(S ∪ P)?
You would need to subtract the intersection n(S ∩ P), but here the digit sets are different (4-digit vs 5-digit), so they are disjoint.
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