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Each rearrangement corresponds to a partition of the set of the six positions into a group of size 3 (the positions that get the letter T), a group of size 1 (the position that gets the letter A), and a group of size 2 (the positions that get the letter O). Thus, the desired number is 1·2·3·4·5·6 6! = = 60. 1! 2! 3! 1·1·2·1·2·3 It is instructive to rederive this answer using an alternative argument. ) Let us rewrite TATTOO in the form T1 AT2 T3 O1 O2 pretending for a moment that we are dealing with 6 distinguishable objects.

Each rearrangement corresponds to a partition of the set of the six positions into a group of size 3 (the positions that get the letter T), a group of size 1 (the position that gets the letter A), and a group of size 2 (the positions that get the letter O). Thus, the desired number is 1·2·3·4·5·6 6! = = 60. 1! 2! 3! 1·1·2·1·2·3 It is instructive to rederive this answer using an alternative argument. ) Let us rewrite TATTOO in the form T1 AT2 T3 O1 O2 pretending for a moment that we are dealing with 6 distinguishable objects.

K = 0, 1, . . , n, where for any positive integer i we have i! = 1 · 2 · · · (i − 1) · i, Sec. 6 Counting∗ 41 and, by convention, 0! = 1. An alternative verification is sketched in the theoretical problems. Note that the binomial probabilities p(k) must add to 1, thus showing the binomial formula n k=0 n k p (1 − p)n−k = 1. 23. Grade of service. An internet service provider has installed c modems to serve the needs of a population of n customers. It is estimated that at a given time, each customer will need a connection with probability p, independently of the others.

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