In how many ways can two pool balls be chosen out of sixteen?

In how many ways can two pool balls be chosen out of sixteen?

• A. 120 ways
• B. 240 ways
• C. 90 ways
• D. 256 ways

Answer: (B)

To determine the number of ways to choose two pool balls from a total of sixteen, we can use the concept of combinations, which is used when the order of selection does not matter. The formula for combinations is given by: \[ \binom{n}{r} = \frac{n!}{r!(n - r)!} \] Where: - \( n \) is the total number of items to choose from (in this case, 16 pool balls), - \( r \) is the number of items to choose (in this case, 2). Substituting the values into the formula, we have: \[ \binom{16}{2} = \frac{16!}{2!(16 - 2)!} = \frac{16!}{2! \cdot 14!} \] Here, the \( 16! \) in the numerator and the \( 14! \) in the denominator cancel each other out: \[ \frac{16 \times 15}{2!} = \frac{16 \times 15}{2 \times 1} = \frac{240}{2} = 120 \] Thus, the total number of

To determine the number of ways to choose two pool balls from a total of sixteen, we can use the concept of combinations, which is used when the order of selection does not matter. The formula for combinations is given by:

[

\binom{n}{r} = \frac{n!}{r!(n - r)!}

]

Where:

• ( n ) is the total number of items to choose from (in this case, 16 pool balls),

• ( r ) is the number of items to choose (in this case, 2).

Substituting the values into the formula, we have:

[

\binom{16}{2} = \frac{16!}{2!(16 - 2)!} = \frac{16!}{2! \cdot 14!}

]

Here, the ( 16! ) in the numerator and the ( 14! ) in the denominator cancel each other out:

[

\frac{16 \times 15}{2!} = \frac{16 \times 15}{2 \times 1} = \frac{240}{2} = 120

]

Thus, the total number of