In the 5-number lottery on September 10, 2005, the following numbers were drawn: 4, 16, 22, 48, 88. All five numbers are even, exactly four of them are divisible by 4, three by 8, and two by 16. In how many ways can five different numbers with these properties be selected from the integers ranging from 1 to 90?
Solution. Among the integers from 1 to 90, every second number, which is 45, is divisible by 2. Among the numbers divisible by 2, every second one is also divisible by 4, making their count 22. Continuing this way: the numbers divisible by 8 are 11, and those divisible by 16 are 5.

According to the condition, exactly two of the 5 numbers must be divisible by 16. For this, we have $\binom{5}{2}=10$ possibilities. But this already gives us two numbers divisible by 8. The third one cannot be among the numbers divisible by 16, so the number of possibilities is $11-5=6$. Furthermore, we need one more number divisible by 4 but not by 8. This can be chosen in $22-11=11$ ways. Finally, we need one number divisible by 2 but not by 4. This provides $45-22=23$ more possibilities.

The lottery numbers that meet the conditions can be chosen from the 90 numbers in $10 \cdot 6 \cdot 11 \cdot 23=15180$ ways.