Answer:
Therefore the inspector can perform three tests in 35 different ways.
Step-by-step explanation:
i) It is given that there is a total of seven tests that the inspector can perform.
ii) it is also given that the inspector needs to select or choose three tests only out of the seven.
iii) therefore this is a Combination problem.
iv) therefore the number of combinations or selections possible is given by
[tex]\binom{7}{3} = \frac{7!}{(3!)(7 - 3)!} = \frac{7!}{(3!)(4!)} = \frac{42 \times 120}{6 \times 24} = \frac{42 \times 20}{24} = \frac{21 \times 10}{6} = \frac{7 \times 10}{2} = 35[/tex]
Therefore the inspector can perform three tests in 35 different ways.
