Combinatorics

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High School Math Grades 9-12, Ages 13-18

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High School Math Grades 9-12, Ages 13-18

Grades 9-12, Ages 13-18

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#1 h Apr 18, 2025, 2:28 PM • 1 Y

Y by EntangledElectron99

1)Two players $A$ and $B$ play a series of of $2n$ games. Each game either results in a win or loss for $A$ . Total number of ways in which $A$ can win the series is:
A) $\frac{1}{2} ( 2^{2n} - \binom{2n}{n})$
B) $\frac{1}{2} ( 2^{2n} - 2\cdot(\binom{2n}{n}))$
C) $\frac{1}{2} ( 2^{n} - \binom{2n}{n})$
D) $\frac{1}{2} ( 2^{n} - 2 \cdot \binom{2n}{n})$
2) A person predicts the outcome of 20 cricket matches of his home team. Each match can either result in a win , loss or tie for the home team. The total number of ways in which he can make the predictions so that 10 predictions is correct is equal to:

A) $\binom{20}{10} \times 2^{10}$
B) $\binom{20}{10} \times 3^{10}$
C) $\binom{20}{10} \times 3^{20}$
D) $\binom{20}{10} \times 2^{20}$

Please mention the solutions properly.

This post has been edited 1 time. Last edited by AzSolver257, Apr 19, 2025, 4:54 AM
Reason: Correct typing error

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SomeonecoolLovesMaths

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SomeonecoolLovesMaths

#2 h Apr 18, 2025, 2:48 PM • 1 Y

Y by EntangledElectron99

AzSolver257 wrote:

1)Two players $A$ and $B$ play a series of of $2n$ games. Each game either results in a win or loss for $A$ . Total number of ways in which $A$ can win the series is:
A) $\frac{1}{2} ( 2^{2n} - \binom{2n}{n})$
B) $\frac{1}{2} ( 2^{2n} - 2\cdot(\binom{2n}{n}))$
C) $\frac{1}{2} ( 2^{n} - \binom{2n}{n})$
D) $\frac{1}{2} ( 2^{n} - 2 \cdot \binom{2n}{n})$
2) A person predicts the outcome of 20 cricket matches hof his home team. Each match can either result in a win , loss or tie for the home team. The total number of ways in which he can make the predictions so that 10 predictions is correct is equal to:

A) $\binom{20}{10} \times 2^{10}$
B) $\binom{20}{10} \times 3^{10}$
C) $\binom{20}{10} \times 3^{20}$
D) $\binom{20}{10} \times 2^{20}$

Please mention the solutions properly.

Solution 1

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