Unit recruitment

Suppose that X people (x≥3) participated in the recruitment interview of this unit.

Then the probability of two people being recruited at the same time is Cx-2 superscript 1/Cx superscript 3= 1/70.

That is 6 (x-2)/[x (x-1) (x-2)] =1/70.

That is x(x- 1)=420,

∴(x-2 1)(x+20)=0，

The solution is x = 2 1.