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Alejandra is designing a game of chance. For one part of the game, a player is to randomly choose \(\mathrm{3}\) marbles, without replacement, from a box containing \(\mathrm{B}\) blue marbles, \(\mathrm{R}\) red marbles, and no other marbles. Alejandra correctly determined the positive integers \(\mathrm{B}\) and \(\mathrm{R}\) so that the number of possible selections in which \(\mathrm{2}\) blue marbles and \(\mathrm{1}\) red marble are chosen is twice the number of possible selections in which \(\mathrm{2}\) red marbles and \(\mathrm{1}\) blue marble are chosen.
The positive integers \(\mathrm{B}\) and \(\mathrm{R}\) that Alejandra determined must be such that \(\mathrm{B}\) is the number that is \(\mathrm{1}\) _ the number that is \(\mathrm{2}\) _ \(\mathrm{R}\). Based on the information provided, select for \(\mathrm{1}\) and for \(\mathrm{2}\) the options that create the most accurate statement. Make only two selections, one in each column.
1
2
2 more than
1 more than
1 less than
2 less than
half
twice
We have a box with:
Box visualization:
[BBBBB...B RRRR...R] B blues R reds
We need to find how B relates to R. The question asks us to fill in:
"B is the number that is [1] the number that is [2] R"
\(\mathrm{C(B,2) \times C(R,1) = 2 \times C(R,2) \times C(B,1)}\)
Substituting:
\(\frac{\mathrm{B(B-1)}}{2} \times \mathrm{R} = 2 \times \frac{\mathrm{R(R-1)}}{2} \times \mathrm{B}\)
Simplifying:
\(\frac{\mathrm{B(B-1)R}}{2} = \mathrm{R(R-1)B}\)
Multiplying both sides by 2:
\(\mathrm{B(B-1)R = 2R(R-1)B}\)
Since B, R > 0, we can divide by BR:
\(\mathrm{B-1 = 2(R-1)}\)
Expanding:
\(\mathrm{B-1 = 2R-2}\)
\(\mathrm{B = 2R-1}\)
\(\mathrm{B = 2R - 1}\) means:
Final Answer:
This makes the complete statement: "B is the number that is 1 less than the number that is twice R"
We can verify: If R = 3, then \(\mathrm{B = 2(3) - 1 = 5}\), and indeed 5 is 1 less than 6, which is twice 3.