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Alejandra is designing a game of chance. For one part of the game, a player is to randomly choose \(3\) 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 \(2\) blue marbles and \(1\) red marble are chosen is twice the number of possible selections in which \(1\) blue marble and \(2\) red marbles are chosen.
The positive integers \(\mathrm{B}\) and \(\mathrm{R}\) that Alejandra determined must be such that \(\mathrm{B}\) is the number that is 1 the number that is 2 \(\mathrm{R}\). Based on the information provided, select for 1 and for 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
Let's create a simple diagram showing our marble selection scenario:
Box Contents: [B blue marbles] + [R red marbles] Selection Scenarios: Scenario 1: Choose 2 blue, 1 red → Number of ways = C(B,2) × C(R,1) Scenario 2: Choose 1 blue, 2 red → Number of ways = C(B,1) × C(R,2) Given: Scenario 1 = 2 × Scenario 2
Let's test with B = 3, R = 2:
- Ways to choose 2 blue, 1 red: \(\mathrm{C(3,2)} \times \mathrm{C(2,1)} = 3 \times 2 = 6\)
- Ways to choose 1 blue, 2 red: \(\mathrm{C(3,1)} \times \mathrm{C(2,2)} = 3 \times 1 = 3\)
- Check: \(6 = 2 \times 3\) ✓
The question asks us to complete: "B is the number that is [1] the number that is [2] R"
This is asking for the mathematical relationship between B and R expressed in two parts.
Number of ways to choose 2 blue and 1 red:
\(\mathrm{C(B,2)} \times \mathrm{C(R,1)} = \frac{\mathrm{B(B-1)}}{2} \times \mathrm{R}\)
Number of ways to choose 1 blue and 2 red:
\(\mathrm{C(B,1)} \times \mathrm{C(R,2)} = \mathrm{B} \times \frac{\mathrm{R(R-1)}}{2}\)
Given condition:
\(\frac{\mathrm{B(B-1)}}{2} \times \mathrm{R} = 2 \times \mathrm{B} \times \frac{\mathrm{R(R-1)}}{2}\)
Let's simplify step by step:
\(\frac{\mathrm{B(B-1)}}{2} \times \mathrm{R} = 2 \times \mathrm{B} \times \frac{\mathrm{R(R-1)}}{2}\)
Dividing both sides by \(\frac{\mathrm{BR}}{2}\):
\(\mathrm{B-1} = 2(\mathrm{R-1})\)
Expanding:
\(\mathrm{B-1} = \mathrm{2R-2}\)
Solving for B:
\(\mathrm{B} = \mathrm{2R-1}\)
B equals 2R minus 1, which means:
- B is 1 less than 2R
- B is 1 less than (twice R)
Looking at our relationship \(\mathrm{B} = \mathrm{2R - 1}\):
- Position [1] needs to describe the relationship to "the number that is [2] R"
- "The number that is [2] R" should be 2R (twice R)
- B is 1 less than this number
Therefore:
- Position [1]: "1 less than"
- Position [2]: "twice"
This gives us: "B is the number that is 1 less than the number that is twice R"
Let's verify with our concrete example:
- If \(\mathrm{R} = 2\), then \(\mathrm{B} = 2(2) - 1 = 3\)
- Check: 3 is 1 less than 4, and 4 is twice 2 ✓
Position 1: 1 less than
Position 2: twice
The complete statement reads: "B is the number that is 1 less than the number that is twice R"
This satisfies the condition that the number of ways to select 2 blue and 1 red marble is twice the number of ways to select 1 blue and 2 red marbles.