Worked solution
Fill the most constrained digit position first
How many numbers between 3000 and 5000 can be formed using only the digits 1, 2, 3, 4 and 5 if (i) no digit is repeated, (ii) the digits can be repeated, (iii) no repetition is allowed and the numbers are even?
CJC · Lecture · Example 15
(i) Count four positions without repetition
The thousands digit must be 3 or 4: 2 choices. The hundreds, tens and units then have 4, 3 and 2 remaining choices. Hence .
(ii) Allow repetition in each later position
The thousands digit still has 2 choices. Each later position independently has 5 choices: .
(iii) First digit 3: fill the units position next
The units digit is 2 or 4, giving 2 choices. Fill tens next with 3 remaining choices, then hundreds with 2. Thus . The displayed positional counts are fixed 3, then 2,3,2; the counting stages run units, tens, hundreds.
(iii) First digit 4: the units digit is forced
Without repetition, the units digit must be 2. Tens has 3 choices and hundreds has 2. The displayed row is fixed 4, then 2 choices,3 choices, fixed 2. This contributes .
(iii) Add the disjoint first-digit cases
There are even numbers. Follow the Teacher guidance: positions need not be filled left to right; usually begin with the most constrained position. Here fix the thousands digit to determine the available even units digits, then complete the middle positions.
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Common mistakes
The first digit changes the units choices in part (iii); do not multiply a single units count across both first-digit cases. Neither endpoint 3000 nor 5000 can be made because 0 is unavailable.