![]() For, the number of ways of selecting 2, is the same as the number of ways of leaving 6 behind. We see that 8 C 2, the number of ways of selecting 2 things from 8, is equal to 8 C 6 ( Example 2), the number of ways of selecting 8 minus 2, or 6. The number of combinations is always a whole number.Įxample 3. The numerator and denominator will always have common factors. The numerator has three factors beginning with the upper index 6 and going down. 6 C 3Īgain, both the numerator and denominator have the number of factors indicated by the lower index, which in this case is 3. The order in which you take them does not matter. In how many different ways could you do that? You have 6 pairs of socks, but on your vaction you will take only 3. The lower index k indicates how many of those we are selecting.Įxample 1. The upper index n indicates the number of distinct things. The numerator has 4 factors starting with the upper index and going down, while the denominator is 4. Notice: The numerator and denominator have the same number of factors, 4, which is indicated by the lower index. This is The number of combinations of 10 distinct things taken 4 at a time. Therefore, the number of combinations is equal to the number of permutations divided by 4!. In other words, the numbers of permutations is 4! times the number of combinations. How many combinations, then, are there of 10 things taken 4 at a time?įor each combination, there are 4! permutations. And it does not matter in which order they appear. ![]() We care only that 4 letters- b, e, g, h, say-have been selected. ![]() Say, however, that we do not care about the order. If we have 10 letters abcdefgahij, then we have seen that the number of permutations of them taken 4 at a time, is Table of Contents | Home PERMUTATIONS AND COMBINATIONS ![]()
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