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There are 2 defective screws in a box having 6 screws. In how many ways can 2 non-defective screws be selected from the box?

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2 screws are defective In a box of 6 screws.

∴ Non-defective screws In a box = 6 – 2 = 4

2 non-defective screws out of 4 non-defective screws can be selected

in 4C2 = \(\frac{4×3}{2×1}\) = 6 ways.

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