Answered

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a committee of 6 is to be selected from 6 lawyers and 9 engineers. In how many ways can 2 lawyers and 4 engineers be chosen to form the committee? 3 lawyers and 3 engineers?6 engineers? all lawyers?

Sagot :

Shinalcantaraidn
this problem involves combination.
for 2 lawyers and 4 engrs you will have:
for 2 lawyers taken from 6lawyers
[tex]C = \frac{6(5)}{2!}[/tex]
[tex]C = \frac{6(5)}{2(1)} [/tex]
[tex]C = \frac{30}{2} [/tex]
[tex]C = 15[/tex]
for 9 engrs taken 4
[tex]C = \frac{9(8)(7)(6)}{4!} [/tex]
[tex]C = \frac{9(8)(7)(6)}{4(3)(2)(1)} [/tex]
[tex]C = \frac{3024}{24} [/tex]
[tex]C = 126[/tex]
multiply the two results to find the answer. you'll have:
[tex]C_t = 15(126)[/tex]
[tex]C_t = 1890[/tex]
or in the calculator, you just have to input,
[tex] _{6} C_2 (_9 C _4) = 1890[/tex]
for the next question - 3 lawyers, 3 engrs
[tex]_6 C _3 (_9 C _3) = 1680[/tex]
for 6 engineers:
[tex]_9 C _6 = 84[/tex]
all lawyers:
[tex]_6 C _6 = 1[/tex]
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