MHB How many ways can equal number of men and women line up for a photo?

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The discussion focuses on calculating the number of ways to line up an equal number of men and women for a photo from a group of 4 men and 3 women. The proposed solution involves using combinations and factorials to account for different group sizes: one man and one woman, two men and two women, and three men and three women. The calculations provided are correct, yielding the total number of arrangements for each scenario. The final answer confirms the validity of the approach taken. This method effectively demonstrates how to solve combinatorial problems involving equal group sizes.
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In how many ways can a non-empty collection of people to be chosen from 4 men and 3 women to lineup for a photo if the number of men must be the same as the number of women?

I said

[math]{4 \choose 1} {3 \choose 1} 2! +{4 \choose 2} {3 \choose 2} 4! + {4 \choose 3} {3 \choose 3} 6![/math]

Is this right?
 
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find_the_fun said:
In how many ways can a non-empty collection of people to be chosen from 4 men and 3 women to lineup for a photo if the number of men must be the same as the number of women?

I said

[math]{4 \choose 1} {3 \choose 1} 2! +{4 \choose 2} {3 \choose 2} 4! + {4 \choose 3} {3 \choose 3} 6![/math]

Is this right?
Yes.
 

Thread 'How do E[X] and E[|X|] relate?'

Greetings, I am studying probability theory [non-measure theory] from a textbook. I stumbled to the topic stating that Cauchy Distribution has no moments. It was not proved, and I tried working it via direct calculation of the improper integral of E[X^n] for the case n=1. Anyhow, I wanted to generalize this without success. I stumbled upon this thread here: https://www.physicsforums.com/threads/how-to-prove-the-cauchy-distribution-has-no-moments.992416/ I really enjoyed the proof...
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