#Help with question on Topic Permutation and Combination

31 messages · Page 1 of 1 (latest)

orchid quartz
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Hey can someone please solve this question, the 3rd part actually, i don't know the answer and can't find the same thing on internet.

My answer is coming out to be 12C4(total selections of 4 out of 12) - 6C2 (selection of 2 couples out of 6) - 6C1(selection of 1 couple out of 6) = 495 - 15- 6 = 474

Idk if its correct so please do rectify if im doin some mistake as im quiet new to the topic.

Thanks : D

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shut canopyBOT
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harsh orbit
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I would solve this using permutation
we can choose anyone first
then we can’t choose that persons partner so 10
then we can’t chose 4 people so we choose from the remaining 8

harsh orbit
orchid quartz
harsh orbit
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because our first choice has a partner

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we can’t chose that partner

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same with #2

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also keep in mind you have to divide that by 6 because combination

orchid quartz
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ohw

blazing olive
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i'd say 240

orchid quartz
blazing olive
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anyone we selcet is part of a couple, so we select 4 couples
6c4=15
and then select 1 of 2 four times

rare elk
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If you don't want to form a couple then you choose 4 people so that you either get zero or one person from each couple. So you choose 4 couples out of the 6 and from each of them one of the two persons. $\binom{6}{4} \cdot (\binom{2}{1})^{4} = 240$

hollow obsidianBOT
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Ludwig

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harsh orbit
orchid quartz
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ya

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got it

rare elk
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you are welcome

orchid quartz
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can i do it by that complimentary count path?

harsh orbit
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yes but it’s harder

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wait you could just take 12C4 and subtract your answers for q1 and q2

orchid quartz
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oh ya 💀

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hmm so i guess my "approach was right" just that i messed up the total ways for counting the couples

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got it all

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thanks everyone

haughty copperBOT
orchid quartz
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uh?

harsh orbit
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+close