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- https://math.libretexts.org/Courses/Chabot_College/Math_in_Society_(Zhang)/08%3A_Counting/8.01%3A_The_Fundamental_Counting_PrincipleIf there are n1 ways to of choosing the first item, n2 ways of choosing the second item after the first item is chosen, n3 ways of choosing the third item after the first two have been ...If there are n1 ways to of choosing the first item, n2 ways of choosing the second item after the first item is chosen, n3 ways of choosing the third item after the first two have been chosen, and so on until there are nk ways of choosing the last item after the earlier choices, then the total number of choices overall is given by
- https://math.libretexts.org/Courses/Fullerton_College/Math_100%3A_Liberal_Arts_Math_(Claassen_and_Ikeda)/06%3A_Probability/6.04%3A_Counting_MethodsSo far the problems we have looked at had rather small total number of outcomes. We could easily count the number of elements in the sample space. If there are a large number of elements in the sample...So far the problems we have looked at had rather small total number of outcomes. We could easily count the number of elements in the sample space. If there are a large number of elements in the sample space we can use counting techniques such as permutations or combinations to count them.
- https://math.libretexts.org/Courses/Las_Positas_College/Math_for_Liberal_Arts/12%3A_Counting/12.01%3A_The_Fundamental_Counting_PrincipleIf there are n1 ways to of choosing the first item, n2 ways of choosing the second item after the first item is chosen, n3 ways of choosing the third item after the first two have been ...If there are n1 ways to of choosing the first item, n2 ways of choosing the second item after the first item is chosen, n3 ways of choosing the third item after the first two have been chosen, and so on until there are nk ways of choosing the last item after the earlier choices, then the total number of choices overall is given by
- https://math.libretexts.org/Courses/Chabot_College/Math_in_Society_(Zhang)/08%3A_Counting/8.02%3A_Permutations_and_Combinations\[\begin{align*} C(13,3) &=\frac{13 !}{3 !(13-3) !}=\frac{13 !}{3 ! 10 !} \\[4pt] &= \dfrac{13 \times 12 \times 11 \times 10 \times 9\times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \tim...C(13,3)=13!3!(13−3)!=13!3!10!=13×12×11×10×9×8×7×6×5×4×3×2×1(3×2×1)(10×9×8×7×6×5×4×3×2×1)=13×12×113×2×1=17166=286
- https://math.libretexts.org/Courses/Mt._San_Jacinto_College/Ideas_of_Mathematics/05%3A_Probability/5.04%3A_Counting_MethodsSo far the problems we have looked at had rather small total number of outcomes. We could easily count the number of elements in the sample space. If there are a large number of elements in the sample...So far the problems we have looked at had rather small total number of outcomes. We could easily count the number of elements in the sample space. If there are a large number of elements in the sample space we can use counting techniques such as permutations or combinations to count them.
- https://math.libretexts.org/Bookshelves/Applied_Mathematics/Book%3A_College_Mathematics_for_Everyday_Life_(Inigo_et_al)/03%3A_Probability/3.05%3A_Counting_MethodsSo far the problems we have looked at had rather small total number of outcomes. We could easily count the number of elements in the sample space. If there are a large number of elements in the sample...So far the problems we have looked at had rather small total number of outcomes. We could easily count the number of elements in the sample space. If there are a large number of elements in the sample space we can use counting techniques such as permutations or combinations to count them.
- https://math.libretexts.org/Courses/Las_Positas_College/Math_for_Liberal_Arts/12%3A_Counting/12.02%3A_Permutations_and_Combinations\[\begin{align*} C(13,3) &=\frac{13 !}{3 !(13-3) !}=\frac{13 !}{3 ! 10 !} \\[4pt] &= \dfrac{13 \times 12 \times 11 \times 10 \times 9\times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \tim...C(13,3)=13!3!(13−3)!=13!3!10!=13×12×11×10×9×8×7×6×5×4×3×2×1(3×2×1)(10×9×8×7×6×5×4×3×2×1)=13×12×113×2×1=17166=286
