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ps://www-awu.aleks.com/alekscgi/x/isl.exe/1o_u-IgNslkr/j8P3jH-liJQhIJODya094wj2gJSklb9agPtGFIZ2O PROBABILITY AND STATISTICSIntroduction to permutations and combinationsSuppose we want to choose 2 objects, without replacement, from the 5 objects pencil, eraser, desk, chair, and lamp.(a) How many ways can this be done, if the order of the choices matters?0(b) How many ways can this be done, if the order of the choices does not matter?I need help with this math problem.

Respuesta :

So, permutation is when the order does matter

The combination is when it doesn't.

[tex]P=\frac{n!}{(r\text{ - }n)!}=\frac{4!}{2!}=\frac{4(3)(2!)}{2!}=4\text{ x }3=\text{ }12[/tex]

That would be a, then: Combination

[tex]C=\frac{n!}{r!(n\text{ - }r)!}=\frac{4!}{2!(2!)}=\frac{4(3)(2!)}{2!(2!)}=\frac{4(3)}{(2)(1)}=\frac{12}{2}=6[/tex]

So, a is 12 and b is 6

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