A box contains 8 two-inch screws. Four have a Phillips head and 4 have a slotted head. In how many ways can 4 screws be chosen so that 2 have a Phillips head and 2 have a slotted head?

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Answer:

There are [tex]6\cdot 6=36[/tex] different ways to choose 4 screws such that 2 have a Phillips head and 2 have a slotted head.

Step-by-step explanation:

If 4 screws must be chosen so that 2 have a Phillips head and 2 have a slotted head, then you have to choose 2 screws with a Phillips head from 4 screws with a Phillips head and 2 screws with a slotted head from 4 screws with a slotted head.

You can choose 2 screws with a Phillips head from 4 screws with a Phillips head in

[tex]C^4_2=\dfrac{4!}{2!(4-2)!}=\dfrac{4!}{2!\cdot2!}=\dfrac{1\cdot2\cdot3\cdot4}{1\cdot2\cdot1\cdot2}=6[/tex]

different ways.

You can choose 2 screws with a slotted head from 4 screws with a slotted head in

[tex]C^4_2=\dfrac{4!}{2!(4-2)!}=\dfrac{4!}{2!\cdot2!}=\dfrac{1\cdot2\cdot3\cdot4}{1\cdot2\cdot1\cdot2}=6[/tex]

different ways.

In total there are [tex]6\cdot 6=36[/tex] different ways to choose 4 screws such that 2 have a Phillips head and 2 have a slotted head.