Maths Olympiad Prep

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Combinatorics Difficulty 6.5 National olympiad Find the answer

Let NN be the number of ways of choosing a subset of 55 distinct numbers from the set
10a+b:1a5,1b5{10a+b:1\leq a\leq 5, 1\leq b\leq 5}
where a,ba,b are integers, such that no two of the selected numbers have the same units digits and no two have the same tens digit. What is the remainder when NN is divided by 7373?

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

1. Understanding the Set:
The set given is {10a+b:1a5,1b5}\{10a + b : 1 \leq a \leq 5, 1 \leq b \leq 5\}. This set consists of numbers formed by 10a+b10a + b where aa and bb are integers from 1 to 5. This results in a 5x5 grid of numbers:
11121314152122232425313233343541424344455152535455 \begin{array}{ccccc} 11 & 12 & 13 & 14 & 15 \\ 21 & 22 & 23 & 24 & 25 \\ 31 & 32 & 33 & 34 & 35 \\ 41 & 42 & 43 & 44 & 45 \\ 51 & 52 & 53 & 54 & 55 \\ \end{array}

2. Choosing 5 Numbers:
We need to choose 5 distinct numbers such that no two numbers have the same units digit and no two numbers have the same tens digit. This means we are essentially choosing one number from each row and each column of the 5x5 grid.

3. Counting the Number of Ways:
The problem reduces to counting the number of ways to select one number from each row and each column of a 5x5 grid. This is equivalent to finding the number of permutations of 5 elements, which is 5!5!.

4. **Calculating 5!5!**:
5!=5×4×3×2×1=120 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120

5. Finding the Remainder:
We need to find the remainder when 120120 is divided by 7373:
120÷73=1remainder47 120 \div 73 = 1 \quad \text{remainder} \quad 47
Therefore, the remainder is 4747.

The final answer is 47\boxed{47}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.