By Klin M., et al.

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Solution. 325. 106. (a) How many different outcomes are possible by tossing 10 similar coins ? (b) How many different outcomes are possible from tossing 10 similar dice ? (c) How many ways can 20 similar books be placed on 5 different shelves ? (d) Out of a large supply of pennies, nickels, dimes, and quarter, in how many ways can 10 coins be selected ? (e) How many ways are there to fill a box with a dozen dough nuts chosen from 8 different varieties of dough nuts ? Solution. (a) This is the same as placing 10 similar balls into two boxes labeled “heads” and “tails”.

7 . 6 . 5 . 4 . 3 . 1 = 3! 3 . 1 We read n ! as “n factorial”. It is true that 4 ! = 24 and 6 ! = 720 but frequently we leave our answers in factorial form rather than evaluating the factorials. Nevertheless, the relation n ! ] enables us to compute the values of n ! for small n fairly quickly. For example : 0 ! = 1, 1 ! =2 3 ! = 6, 4 ! = 24, 5 ! = 120 6 ! = 720, 7 ! = 5040, 8 ! = 40320 9 ! = 362880, 10 ! = 3628800, 11 ! = 39916800. 10. Enumerating r-permutations without repetitions P(n, r) = n(n – 1) ......

For example, A1A2A3BN1N2 and A1A2A3BN2N1. 18 COMBINATORICS AND GRAPH THEORY In fact, the 720 permutations can be listed in pairs whose members differ only in the order of the two N’s. This means that if the tags are dropped from the N’s only 720 or 360 distinguishable permu2 tations remain. Reasoning in a similar way we see that these can be gruoped in groups of 3 ! or 6 that differ only in the order of the three A’s. For example, one group of 6 consists of BNNA1A2A3, BNNA1A3A2, BNNA2A1A3, BNNA2A3A1, BNNA3A1A2, BNNA3A2A1.

### Algebraic combinatorics in mathematical chemistry by Klin M., et al.

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