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Combinatory, Probability Combinatory Permutation: in how many ways we can arrange n elements P=n!=1*234n 1!=1 0!=1 Proof by total induction: n=1: 1!=1 possibility: a n=2: 2!=2 possibilities: ab, ba n=3: 3!=6 possibilities: abc, acb, cab, bac, bca, cba Now, we have to prove that if the statement stands for n=k, then it is true for n=k+1 as well Let’s take n elements. We can arrange them in n! ways If we add one more element, this element can fit into each order in n+1 ways. Therefore n!*(n+1)=(n+1)! Repeated permutation: in how many ways can we arrange n elements if there are identical elements as well P i =n!/(k 1 !*k 2 !k 3 !k n !) Variation: choosing k elements from n elements (order is important) V n k=n*(n-1)(n-2)(n-(k-1))=n!/(n-k)! Repeated variation: choosing k elements from n elements (order is important and we can choose one element more than once) V in k=nk Combination: choosing k elements from n elements (order is not important) We can choose k elements from n elements

n!/(n-k)! times, but since order is not important, we have to divide by k! (we can arrange the chosen element this many ways). C n k=n!/((n-k)!*k!) Repeated combination: choosing k elements from n elements (order is not important and we can choose one element more than once) C in k=C n+k-1 n-1 Theorems in combinatory: Statement: (C n k)= (C n n-k) Proof: n!/((n-k)!*k!)=n!/(k!(n-k)!) Statement: (C n k)+ (C n k+1)= (C n+1 k+1) Proof by examining the two cases in which the extra element is either chosen or not Binomial coefficients: 1 11 121 1331 14641 n=0 n=1 n=2 n=3 n=4 (sum of the coefficients) 1 2 4 8 16 Binomial theorem: (a+b)n=( 0 n)anb0+( 1 n)an-1b1++( n n)a0bn Statement: ( 0 n)+( 1 n)+( 2 n)++( n n)=2n Proof: (1+1)n= we use the binomial theorem Statement: ( 0 n)-( 1 n)+( 2 n)-+/- ( n n)=0 Proof: (1-1)n= we use the binomial theorem Probability p(A)= good events / all events p(A)+p(Ā)=1 1st axiom: 0≤p(a)≤1 2nd axiom: p(certain event)=1; p(impossible event)=0 3rd axiom:

p(AUB)=p(A)+p(B)-p(A∩B)  p(AUB)=p(A)+p(B) if p(A∩B)=0. The events are independent if and only if p(A∩B) = p(A)*p(B). Conditional probability of an event: p(A│B) = p(A∩B)/p(B) Total probability of an event: p(A) = ∑p(A│B k )*p(B k ) = ∑p(A∩B k ) Binomial distribution of an event: p(A)=( k n)*p(A)kp(Ā)n-k Expected value: E(x)=∑x i *p i Application of combinatory and probability - weather forecast - infrastructure: installing traffic lights, bus stops, etc. - economy: insurance companies: risk calculation banks: risk calculation and interest determination stock markets: expected price of shares - games: pools, lottery, card games - biology: dominant and regressive properties - demography: life-expectancy - mathematics: binomial theorem (for bracket expansion)