In the first round, one person chooses 3 people. In round 2, those 3 people choose 3 people each. Keeping with the pattern, how many people get chosen in round 6?

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Answer 1 · 2017-02-06T10:29:52

$3^{6} = 729$ $3^6=729$

We have six Rounds of choosing people:

R1: 3 are chosen

R2: 3 choose 3, or $3 \times 3 = 3^{2} = 9$ $3xx3=3^2=9$ are chosen

R3: 9 choose 3, or $9 \times 3 = 3^{2} \times 3 = 3^{3} = 27$ $9xx3=3^2xx3=3^3=27$ are chosen

and so on.

And so we can conclude that the pattern of the number of people being chosen in any given round is:

$3^{R}$ $3^R$ , where $R = round$ $R="round"$

Therefore, in the 6th round, we'll have:

$3^{6} = 729$ $3^6=729$