Question

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?
PLEASE ANSWER SOON BECAUSE DUE IN 1 HOUR. A LOT OF POINTS!

Answer 1

Answer:

There are $6\cdot 6=36$ 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

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

different ways.

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

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

different ways.

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