Can someone please help me check whether my reasoning is correct for the following two questions? There were similar questions in the forum (1, 2) but they seem incorrect or I don’t understand what they’re doing. (These questions are inspired by a similar one from Magoosh.)
How many different three-letter “words” can be made from the nine letters in ABCDEFGGG? Different arrangements of the same three letters (such as ABC and CBA) count as different words. However, ABG_1 and ABG_2 count as one word.
First, find the number of cases where there are no repetitions. So only using ABCDEFG, it’ll be 7 \times 6 \times 5 = 210.
Second, find number of cases where there are two repetitions. In terms of arrangements, there are: \_GG, G\_G, and GG\_. Since there are ABCDEF to fill in the slot, so each arrangement is 6\times1\times1=6. And because there are three arrangements, it’ll be 6\times3=18.
Since there are three G's, why not 6\times3\times2 instead of 6\times1\times1? Because we’ll count repeats if we do that.
Third, find the number of cases where there are three repetitions. In terms of arrangements, there is only GGG, which is just 1.
Finally, sum the number of cases and you have 210+18+1=229.
How many different three-letter “words” can be made from the six letters in BANANA? Different arrangements of the same three letters (such as BAN and NBA) count as different words. However, BA_1N_1 and BA_1N_2 count as one word.
First, find the number of cases where there are no repetitions. So only using BAN, it’ll be 3\times2\times1=6
Second, find the number cases where there are two repetitions for N. In terms of arrangements, there are: \_NN, N\_N, and NN\_. Since there are BA to fill in the slot, so each arrangement is 2\times1\times1. And because there are three arrangements, it’ll be 2\times3=6.
Third, find the number cases where there are two repetitions for A. In terms of arrangements, there are: \_AA, A\_A, and AA\_. Since there are BN to fill in the slot, so each arrangement is 2\times1\times1. And because there are three arrangements, it’ll be 2\times3=6.
Fourth, find the number of cases where there are three repetitions for A. In terms of arrangements, there is only AAA, which is just 1.
Finally, sum the number of cases and you have 6+6+6+1=19.