Иван Трещев – Discrete Math. Practice. For students of technical specialties (страница 3)
Answer:142729.
Task number 22: preference. Cards are laid out in 3 piles and 2 cards are placed in the draw. Play 32 cards, i.e. each player receives 10 cards. Determine the number of layout options.
Solution: using the formula of permutations with repetitions we get: ((32!) \ (10!3*2!))
Answer: ((32!) \ (10!3*2!)).
Task number 23: from a group of 15 people it is necessary to select a team, which should include at least 5 people. How many choices.
Solution: we calculate the number of unfavorable combinations of choice, i.e., we compose options for brigades of 1, 2, 3, 4 people. Their number is: C151+ Cn152+C153+ C154=1940.
And the total number of brigades is 215. The difference gives the number of noble combinations.
Answer: 215 – 1914.
Task number 24: three boys picked 10 apples. How many ways are there to split apples between them.
Solution: We write 10 units and 2 zeros that perform the functions of a separator, and then we begin to rearrange them with all possiblemethods. Each permutation will correspond to some way of dividing 10 apples into 3 heaps. Each section method will correspond to some code containing 10 units and 2 zeros. Therefore, the number of partition methods: P (10.2) = ((10!) \ (2!8!)) =45
Answer: 45.
Problem number 25: a set of 8 elements is necessary to find the number of its partitions.
Solution: The number of partitions of the set is the Bell number => B8=4140.
Answer: 4140.
Problem number 26: find the number of decimal non-negative integers, consisting of numbers, arranged in increasing order.
Solution: 1+ C91+…C99=210.
Answer: 210
Problem number 27: how many word decompositions from the letters ABRAKADABRA
Solution: in the word ABRAKADABRA there are 11 letters, of which five are letters A, two letters B and two letters R. So, we get by the formula of permutations with repetitions, we get: Pn = ((11!) \ (5!*2! *2!)) =83160
Answer: 83160.
Problem number 28: find the sum of all divisors of 300.
Solution: 300 = 1+2+3+4+5+6+10+12+15+20+25+30+50+60+75 +100+150 +300 = 868
Answer: 868. This problem can be solved using the Euler formula.
Problem number 29: How many numbers between 1000 and 10000 consist of:
– odd numbers;
– different numbers.
Solution:
a) Between 1000 and 10000 are four-digit numbers. Because numbers consist only of odd numbers, then the first digit can be selected in 5 ways (digits 1, 3, 5, 7 and 9), the second, third and fourth digit can also be selected in 5 ways (digits 1, 3, 5, 7 and 9). Therefore, there will be a total of such numbers: n=5*5 *5 *5=625
b) numbers consist of various digits, then the first digit can be selected in 9 ways (digits 1, 2, 3, 4, 5, 6, 7, 8 and 9), the second digit in 9 ways (there can be 0 and any of the previous ones, but not repeating), the third digit in 8 ways and the fourth digit in 7 ways. Therefore, there will be a total of such numbers: n=9*9 *8 *7=4536
Answer: a) 625; b) 4536.
The task №30: prove | √ |x||=√x||.
Solution: Let m = | √ |x||
Because the rules ≤ |x|=n <=> nx <n1 should m ≤ √ |x| <m+1
m2 ≤ |x| (<m+1) 2
From the rules: x <n <=> |x| <n
n ≤x <=> n <|x|,where nan integer
must be m2 ≤ x (<m+1) 2
m ≤ √ x <m+1
From (*) => m= | √
Thus | √ |x||= m= | √x|.
Problem No. 31: un +2 = -un +1 +2un, u0 = 0, u1 = 1
Solution: here K (x) =1+x-2x2. We calculate: D (x) =K (x) u (x) = (1+x-2x2) (u0 + u1x+…) 0+x+0=x
We get: u (x) = ((x) \ (1+x-2x2))
The next step is the expansion of the denominator
and
a1=1
a1= ((-1) \ (2))
arrive to the formula: u (x) = ((x) \ (12+x-x2)) = ((x) \ ((1-x) (1+ ((x) \ (2)))))
We find the decomposition into a sum of simple fractions by the method of indefinite coefficients: ((x) \ ((1-x) (1+ ((x) \ (2))))) = ((A) \ ((1-x))) + ((B) \ (1+ ((x) \ (2))))
We obtain a system of linear equations: ((1) \ (2)) AB=1B+A=0
Its solution: A= ((2) \ (3)),B= ((-2) \ (3)). Hence: u (x) = (((2) \ (3)) \ (1-x)) + (((-2) \ (3)) \ (1- ((x) \ (2)))) = ((2) \ (3)) ∑k=0∞1kxk– ((2) \ (3)) ∑k=0∞ ((-1) \ (3)) kxk
This leads to the answer: un= ((2) \ (3)) – ((2) \ (3)) * ((-1) \ (2)) n
Answer: un= ((2) \ (3)) – ((2) \ (3)) * ((-1) \ (2)) n
Task number 32: 15 students shook hands at a meeting of students, three people made 4 handshakes, and others – 3. How many students were there.
Solution: We will consider a case requiring a smaller number of participants. Since the three made 4 handshakes: we consider. That between them they made maximum handshakes – two, and formed a complete 3x vertex graph and used only 3 handshakes out of the total:
Each «had» two handshakes. The minimum number of vertices that must be added so that the condition «three made 4 handshakes» is met – these are two. Let’s portray them like this:
We get already 9 used handshakes. And the two added vertices have 3 handshakes, which corresponds to the condition. It remains 6. Just the full 4-vertex graph gives us 6 handshakes for each of the 4 students added:
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