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15 Dec 2022, 09:59
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Hi all, would be extremely grateful if there's a Math Expert who might be able to answer this query (perhaps Bunuel as I know you posted the two following examples that I will provide)
My question surrounds the following: suppose that you've already determined whether the question is looking for permutations or combinations as the correct answer. There now appear two contradictory cases on factorial adjustments, attached below:
Case 1: In how many ways can one divide 12 different chocolate bars into four stacks of 3 bars each?
Solution: Dividing equally means forming 4 groups of 3. We don't care about order within each stack. Ordering the 12 bars gives 12! arrangements, and then answer to Q might first look like \(\frac{12!}{3!\times 3!\times 3!\times 3!}\). BUT the stacks are not distinct. When the question says that you need to make n groups/bundles/teams that are not distinct, you need to divide by (n!). If the groups/bundles/teams are distinct then you do not divide by (n!).
Case 2: How many different 6-digit positive integers are there, where 3 of the digits are each one of the digits 5 or 7, and the other 3 digits are each of one of the digits 1, 4, 6, or 8 ?
Solution: The 3 digits coming from \({5,7}\) can be selected and arranged in \(2\times 2\times 2\) ways, and similarly the 3 digits from \({1,4,6,8}\) can be selected and arranged in \(4\times 4\times 4\) ways. Now, we need to mix up these 2 sets of 3 digits by applying a factorial adjustment. The total number of ways is then = \(2^3 \times 4^3 \times \frac{6!}{3!3!}\), giving \(10,240\) total ways.
Now, as we can see from the bolded section of the Case 1 solution, the adjustment following a string of multiplications is either to divide (when ordering does not matter), or leave things untouched (if ordering does matter). Intuitively, this is because the string of multiplications would represent arranging, right? And then if you don't care about arrangement (i.e. you've determined that the question is looking for the number of combinations), you would divide by the relevant factorial adjustment. BUT in Case 2, you're multiplying a string of multiplications by a factorial adjustment - whilst the bolded advice from Case 1's solution tells us you only either divide, or leave the string of multiplications untouched (and therefore never a case where you have to multiply the string of multiplications by a further factorial adjustment).
Put differently, suppose I've determined that the question is asking me to look for permutations. I'm then quite confused over deciding when I need to multiply by a factorial correction, and when I should leave things untouched because my string of multiplications already incorporates the factorial adjustments. It appears, at first glance, that Case 1 and Case 2 are quite contradictory to me and I'm quite curious to see how this can be reconciled!
Thank you so much!
- FictionFire
My question surrounds the following: suppose that you've already determined whether the question is looking for permutations or combinations as the correct answer. There now appear two contradictory cases on factorial adjustments, attached below:
Case 1: In how many ways can one divide 12 different chocolate bars into four stacks of 3 bars each?
Solution: Dividing equally means forming 4 groups of 3. We don't care about order within each stack. Ordering the 12 bars gives 12! arrangements, and then answer to Q might first look like \(\frac{12!}{3!\times 3!\times 3!\times 3!}\). BUT the stacks are not distinct. When the question says that you need to make n groups/bundles/teams that are not distinct, you need to divide by (n!). If the groups/bundles/teams are distinct then you do not divide by (n!).
Case 2: How many different 6-digit positive integers are there, where 3 of the digits are each one of the digits 5 or 7, and the other 3 digits are each of one of the digits 1, 4, 6, or 8 ?
Solution: The 3 digits coming from \({5,7}\) can be selected and arranged in \(2\times 2\times 2\) ways, and similarly the 3 digits from \({1,4,6,8}\) can be selected and arranged in \(4\times 4\times 4\) ways. Now, we need to mix up these 2 sets of 3 digits by applying a factorial adjustment. The total number of ways is then = \(2^3 \times 4^3 \times \frac{6!}{3!3!}\), giving \(10,240\) total ways.
Now, as we can see from the bolded section of the Case 1 solution, the adjustment following a string of multiplications is either to divide (when ordering does not matter), or leave things untouched (if ordering does matter). Intuitively, this is because the string of multiplications would represent arranging, right? And then if you don't care about arrangement (i.e. you've determined that the question is looking for the number of combinations), you would divide by the relevant factorial adjustment. BUT in Case 2, you're multiplying a string of multiplications by a factorial adjustment - whilst the bolded advice from Case 1's solution tells us you only either divide, or leave the string of multiplications untouched (and therefore never a case where you have to multiply the string of multiplications by a further factorial adjustment).
Put differently, suppose I've determined that the question is asking me to look for permutations. I'm then quite confused over deciding when I need to multiply by a factorial correction, and when I should leave things untouched because my string of multiplications already incorporates the factorial adjustments. It appears, at first glance, that Case 1 and Case 2 are quite contradictory to me and I'm quite curious to see how this can be reconciled!
Thank you so much!
- FictionFire
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