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Number of Flavours | Combinations of Flavours | |
Part 1 | 3 | abc |
Part 2 | 2 | ab |
" | 2 | bc |
" | 2 | ca |
Part 3 | 1 | a |
" | 1 | b |
" | 1 | c |
Number of flavours | Combination of Flavours | |
Part 1 | 3 | abc |
Part 2 | 2 | ab |
" | 2 | ac |
" | 1 | a |
Part 3 | 2 | bc |
" | 1 | b |
" | 1 | c |
This works almost exactly as the previous structure except now the 5th choice is decided by part 2.
Part 2 has 6 combinations {ab,ca; ca,ab; ab,bc, bc, ab; ca,cb; cb,ca}
Part 3 has 2 combinations, the two different orders for the remaining single flavours of ice cream.
Therefore the total number of permutations for this structure is 6 x 2 = 12
To get the total number of permutations across both structures we have:
36 (the number of permutations for the basic structure) + 12 (the number of permutations for the complex structure) = 48
Well done! Some very thorough examplanations here. We look forward to hearing more from you all.