What constraints must a number satisfy to be strictly between 5000 and 10000 using the digits {1, 3, 5, 7, 9}?
No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.
Correct answer
Option analysis
Believing that only one specific digit can occupy the thousands place, or dividing the final result erroneously. Count all valid choices for the thousands place (5, 7, or 9) and multiply by the permutations of the remaining digits: 3 × (4 × 3 × 2) = 72.
Ignoring the lower bound condition of 5000 and calculating all possible 5-digit permutations (5! = 120) or all 4-digit permutations using any of the 5 digits. Numbers strictly between 5000 and 10000 must be 4-digit numbers starting specifically with 5, 7, or 9, not any digit.
None. This is the correct calculation: 3 options for the thousands place multiplied by 4 × 3 × 2 for the remaining three places. Correctly identified that numbers must be 4 digits with thousands digit ∈ {5, 7, 9}, giving 3 × P(4, 3) = 3 × 24 = 72.
Computing permutations for only the last three digits for a single leading digit and forgetting to multiply by the number of valid leading digits, or computing 3! instead of 3 × P(4, 3). The remaining 3 digits are chosen from 4 available digits, yielding 4 × 3 × 2 = 24 arrangements per valid thousands digit, across 3 possible thousands digits.
Digits available: (5 distinct non-zero digits). No repetition is allowed. The numbers must lie strictly between and .
Find the total count of valid 4-digit numbers strictly between and formed without repeating any digits.
Any number strictly between and must be a 4-digit number. For the number to exceed , its thousands digit must be . From the set , the candidates for the thousands place are or (3 choices). The remaining 3 positions (hundreds, tens, units) are filled by choosing and permuting 3 of the remaining 4 digits.
Number of choices for the thousands digit (can be or ). Number of ways to fill the remaining three places from the remaining 4 digits is . Total valid numbers .
Total 4-digit numbers from 5 distinct digits without repetition is . By symmetry, each of the 5 digits appears equally at the thousands place, i.e., times. The digits are (3 digits), so . Both routes yield 72.
What constraints must a number satisfy to be strictly between 5000 and 10000 using the digits {1, 3, 5, 7, 9}?
It must be a 4-digit number whose thousands digit is 5, 7, or 9.After selecting a valid digit for the thousands place, how many ways can the remaining 3 places be filled from the remaining 4 digits?
4 × 3 × 2 = 24 waysWhat is the total number of valid numbers formed?
3 × 24 = 72Quick checks
No, because the available digits are only {1, 3, 5, 7, 9}, none of which is 0. Any number starting with 5 formed with the remaining digits will strictly exceed 5000.