Exercise · ZW-PERMUTATIONS-AND-COMBINATIONS-004 · ★☆☆ Basic

How many different sets of 6 numbers can be chosen from 49, regardless of order?

Choose one option.

Show answer
  • D. \(13\,983\,816\)

Explanation

\(\binom{49}{6}=13\,983\,816\). The figure 10 068 347 520 also counts orderings.

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Review the concept: Permutations and combinations

A permutation \(P(n,k)\) is an ordered arrangement of \(k\) elements from a set of \(n\); a combination \(\binom nk\) is a selection in which order does not matter.

\[P(n,k)=\frac{n!}{(n-k)!},\quad \binom{n}{k}=\frac{n!}{k!\,(n-k)!}\]

More practice problems: permutations and combinations

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