# GRE quant question (hard)?

## 1 Expert Answer

Arnold V. answered • 09/30/19

Experienced Tutor in Math and Physics with PhD in Physics

We assign four students to 3 sections so that at least one student is in each section. Therefore, two students are placed into one of the sections and one student is placed into each of the other two sections. In other words, we distribute a set of 3 different objects (for example, one pair BJ and two single students P and W) putting 1 object into each one of 3 sections. We know that number of permutations for 3 objects is 3! = 3*2*1 = 6. Choosing a different pair of students from 4 students we have 4!/[2!(4 - 2)!] = 6 different sets, namely, {BJ, BP, BW, JP, JW and PW}. Again, each set has 6 permutations, and thus 6 different sets can be assigned in 6x6 = 36 ways.

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Jill M.

This is known as a permutation. Ways for position one to happen (since there are four people,4), times ways for position two to happen (since now there are only three people left,3), times ways for position three to happen (since now there are only two people left,2). Solve 4 x 3 x 2.04/13/19