Mathematical Thinking Problem-solving And Proofs – K 2 3 4 5 6 7 8 9 10 11 12 probability 361 362 363 364 365 366 365 364 363 362 361
5.∏ In an alphabet of size m, there are ml words of length and in l i= 1 (l+ 1 −i) of them, each letter is used at most once. A letter must be selected for each position in the word. When replay is enabled, there are choices at each of the positions, regardless of previous choices. When repetition is forbidden, the number of ways to fill the ith position is l+ 1 −i, regardless of how the previous positions were filled. In any case, by multiplying these factors, arrangements with the product rule are calculated.
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Choose this, there are ways to choose image 2. In general, for every way of choosing images 1 ,.. ., you have, there are ways to choose image + 1. By the product rule, the number of ways to create a bijection is
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∏n− 1 i= 0 (n−i). 5. There are 12 · 47 + 1 · 48 ways to pick two cards from a standard deck of 52 cards such that the first card is a spade and the second card is not an ace. There are 13 ways to start with spades. If there is no ace of spades, there are 47 ways to choose the second card because a non-ace was used. If there is an ace of spades, there are 48 ways to choose the second card. Combining these two cases results in the answer 12· 47 + 48 Alternatively, spades can be called for the first card and no aces for the second card, thereby eliminating cases where the same card is called twice. The result is 13·48 −12, which is equal to the above value.
, according to the binomial theorem. The value is 94 ·· 38 ·· 72 ·· 61, which equals 126.
, the total number of possible hands. a) Hands that have at least three cards of the same rank, we simply pick three cards of the same rank and then pick two more cards, we can get four of the same rank; such hands would count four times. So we count both cases separately. By choosing a rank and another card, there are 13 · 48 hands with four in one rank. In the second case, we choose a rank, omit one of this rank and select two cards of another rank, in 13· 4 ·
B) Hands that have at least two cards of the same value and the numerator from part (a), we can add hands that have two but not three of the same value. Note that we must avoid double counting hands that have a pair from each of the two levels. Alternatively, we can subtract from the total number of hands that have no pair of cards of the same rank. Since we are choosing five different ranks and one card from each, there is
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52 · 48 · 44 · 40 · 36 52 · 51 · 50 · 49 · 48 =. 5.10 2 ncoins toss, the probability of getting exactly lynheads is
/ 22 n There are 2 2 lists of heads and tails, all equally likely. The list of heads is determined by selecting the location of the heads in the list. I mean
Havenheads results. When n=10, the value is ( 19 · 17 · 13 · 11 )/ 218 after cancellation. This equals approx.
54 −k/ 64. k 0 1 2 3 4 instances 625 500 150 20 1 probability .4823 .3858 .1157 .0154.
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5.14 from the sum of three selections z[n]is(n− 1 )(n− 2 )/( 2 n 3 ). There are 3 equally likely outcomes. The number of outcomes that add up to ni is the number of positive integer solutions tox 1 +x 2 +x 3 = nin. Evidence 1 (selections with repetition). The number of solutions is the number of solutions toy 1 +y 2 +y 3 =n−3 in non-negative integers. This is equal to the number of selections of n−3 elements of three types with repetition allowed, i.e
, according to Theorem 5. The probability is therefore (n− 1 )(n− 2 )/( 2 n 3 ). Exhibit 2 (Summary). When x 1 =i, there are i-1 ways to assign the last two values, since x 2 can take any value from 1 ton-i-1, which determines x 3. So the number of solutions is
∑n− 2 i= 1 (n−i− 1 ). The sums are integers from 1 ton−2 (in reverse order), so the sum is (n− 1 )(n− 2 )/2.
5.15 the size of the union of finite sets disjoint in pairsA 1 ,.. ., is the sum of their sizes. We use induction cookers. Basic step: k=1. The set A 1 is also a union and its size is its size. Induction stage: k>1. Let B be the union of A 1 ,.. ., Ak− 1. According to the induction hypothesis |B|=
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∑k− 1 i= 1 You have. Since Ak is disjoint from each of A 1 ,.. ., Ak− 1 , also Ak∩B=∅. Now Corollary 4 states that|Ak∪B|=
∏k i= 1 ri. Basic step: k=1. Elements Tare ways of performing the step, so|T|=r 1. Induction phase:k>1. We divide into sets according to how the first k-1 steps are performed. The given condition means that every set has a sizerk. The induction hypothesis assumes that there is
∏k− 1 i= 1 rises in the partition. Since each has a sizerk and the sets are pairwise disjoint, the sum rule implies that |T|=
5.17 the only solution of luin!+m!=k!in positive integers is n=m= 1 and k= 2. Suppose that n!+m!=k!; by symmetry we can assume that n≥m. Since!>0, we have!>n!≥m!. Using the definition of factorial, we divide the equation n! to get 1+m!/n!=k(k− 1 )· · ·(n+ 1 ). Since 1 and k(k− 1 )· · ·(n+ 1 ) are integers, m!/n! must be an integer. Because we need this. Now we have 2=k(k− 1 )· · ·(n+ 1 ). This requires n+ 1 ≤ 2 and k=n+1, leaving only the option (n, m, k)=( 1, 1, 2). This option is really a solution.
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When k=6, we arrange the six elements of the set of size 10, so the number is 10·9·8·7·6·5 =151200. When k=5, we pick the repeating digit, pick its two positions, and arrange four of the remaining digits into the remaining positions. The number is 10
. We list the number of hands and rank for each distribution. To count hands, we first assign multiples to the suits and then select the specified number of cards from each suit. The number of ways multiples can be assigned depends on how many times each multiplication occurs. With four different multiplicities, there are 24 ways to match them to suits. When three numbers appear (one is repeated), as in 5440, there are 12 ways. With three suits of the same multiplicity as in 4333, there are 4 ways (that’s why this split is so low). Since 13 is odd, there cannot be four suits of the same multiplicity or two pairs of the same multiplicity.
=k!(nn−!k)!. The formula holds for forn=0 by convention that the “factorial” of a negative number is infinite. For n>1, we use Pascal’s formula and the induction hypothesis to obtain (n k
X 0 y 0. Now suppose that the exponent holds when the exponent is n. We consider the sum when the parameter is n+1. The induction hypothesis tells us that (x+y)n=
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Xkyn−k. Since we want the expansion for (x+y)n+ 1 , we multiply both sides of (x+y). To simplify the resulting expression, we want to combine terms in which the exponents onxagree and onyagree. Therefore, we move the index in the first sum. Then we use Pascal’s formula to combine the corresponding terms in the two summations. For terms that do not match, we have
This proof is a direct generalization of the calculation that, for example, obtains the expansion of (x+y) 3 from the expansion of (x+y) 2 .
Xk From the binomial theorem, we can calculate sets of positive even size and sets of negative odd size by setting x= −1. The sum value becomes the total number of even subsets minus the total number of odd subsets. Setting ax= −1 on both sides gives
(−1)k=(1−1)n=0. Thus, the number of subsets of each type is the same. When n=0, there is an even subset and no odd subset, leading to the convention in combinatorics that 0 0 =1.
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Solution in non-negative integers. Exhibit 1 (Summary). We calculated the solution tox 1 + · · · +xk=m for each m, the number is
Solution. Proof 2 (transformation). Introduce another variablexk+ 1. The required solutions correspond to non-negative integer solutions
∑k i= 1 yi=n−k because subtracting 1 turns a positive integer into a non-negative integer, and this is invertible. The formula for repeated selections then says that there is
Such solutions. Proof 2 (direct bijection). In a points and bars pattern, everyone counts points between consecutive bars. If there is at least one point for each xi, then the rods must be placed at indistinct locations between the points. So there are -1 places the bars can go, and we pick -1 of those to determine the positive integer solution.
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Is 1 ifk=1 and 0 ifk>1. So the identity holds when n=1. On induction