Discrete Math. Practice. For students of technical specialties

*6! + C

*5! -C

*4! + C

*3! -C

*2! + C

*1!=142729

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!

*2!))

Answer: ((32!) \ (10!

*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: C

+ C

+C

+ C

=1940.

And the total number of brigades is 2

. The difference gives the number of noble combinations.

Answer: 2

 – 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 => B

=4140.

Answer: 4140.

Problem number 26: find the number of decimal non-negative integers, consisting of numbers, arranged in increasing order.

Solution: 1+ C

+…C

=2

.

Answer: 2

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: P

 = ((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||.