A four letter code has to be formed using the alphabets from the set {a, b, c,…

When a counting problem restricts how many times one particular symbol may appear (here, an odd number of a's), split the total into the individual repetition-count cases that satisfy the restriction. Count each case using the permutations-of-a-multiset rule – arrange n symbols, some repeated, in n!÷(repeated symbol's factorial) ways – then combine the cases with the sum rule, since a code cannot belong to two cases at once.

1. A four-letter code can contain 0, 1, 2, 3, or 4 copies of 'a'; among these only 1 and 3 are odd, so exactly two mutually exclusive cases must be counted and added.

2. Case A – exactly one 'a': the other three positions must be filled by b, c and d, each used exactly once, since they cannot repeat and there are exactly three of them for three slots. This is simply an arrangement of the four distinct symbols a, b, c, d in four positions: 4! = 24 ways.

3. Case B – exactly three 'a's: the remaining one position is filled by a single letter chosen from {b, c, d} – 3C1 = 3 ways to choose it – and the four symbols (three identical a's plus the chosen letter) are then arranged in the four positions. Since the three a's are indistinguishable, this gives 4!÷3! = 4 arrangements for each choice.

4. Multiply the choice count by the arrangement count in Case B (3 × 4 = 12), then add Case A and Case B by the sum rule, since the two cases cannot occur together.

• Case B's count of 4 can be confirmed independently: with three identical a's and one other letter, the code is fully determined by which one of the four positions holds the non-a letter – there are exactly 4 such positions, matching 4!÷3! without needing the factorial formula at all.

• Case A's count is the standard permutation of four distinct symbols and needs no further check.

Adding the two cases: 24 + 12 = 36 different codes can be formed.