Question

Suppose there are ten names in a box and three are to be drawn out. Find the number of ways in which those three names can be taken out. The order of the names does not matter.


7


3


98


120

Answer 1

To find the number of ways to select three names from a box of ten names where the order does not matter, we can use the combination formula:

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

Where:
- \( n \) is the total number of items (in this case, names),
- \( r \) is the number of items to choose.

Here, \( n = 10 \) and \( r = 3 \). Applying the formula:

\[
\binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10!}{3! \times 7!}
\]

Calculating this step by step:

1. Calculate \( 10! \), but we only need the terms down to \( 7! \):
\[
10! = 10 \times 9 \times 8 \times 7!
\]

2. Substitute this back into the equation:
\[
\binom{10}{3} = \frac{10 \times 9 \times 8 \times 7!}{3! \times 7!}
\]

3. The \( 7! \) cancels out:
\[
\binom{10}{3} = \frac{10 \times 9 \times 8}{3!}
\]

4. Calculate \( 3! = 3 \times 2 \times 1 = 6 \):
\[
\binom{10}{3} = \frac{10 \times 9 \times 8}{6}
\]

5. Perform the multiplication:
\[
10 \times 9 = 90
\]
\[
90 \times 8 = 720
\]

6. Finally, divide by 6:
\[
\frac{720}{6} = 120
\]

Thus, the number of ways to draw three names from ten, where order does not matter, is \(\boxed{120}\).