1. **Problem:** How many ways can we select one white and one black square on a chessboard? And how many ways to select two squares of any color? The chessboard has 64 squares, half white (32) and half black (32). - To select one white and one black square: number of ways = $32 \times 32 = 1024$. - To select any two squares: number of ways = $\binom{64}{2} = \frac{64 \times 63}{2} = 2016$. 2. **Problem:** How many ways to select one white and one black square so that they are not in the same row or column? - Total white squares: 32, black squares: 32. - Each row and column has 4 white and 4 black squares. - Total pairs without restriction: $32 \times 32 = 1024$. - Pairs in the same row: There are 8 rows, each with 4 white and 4 black squares, so $8 \times (4 \times 4) = 128$. - Pairs in the same column: Similarly, $8 \times (4 \times 4) = 128$. - Pairs in same row or column: $128 + 128 = 256$. - But pairs in same row and column (same square) counted twice, but impossible since colors differ. - So pairs not in same row or column: $1024 - 256 = 768$. 3. **Problem:** Péter picks either an apple or a peach from a basket with 12 apples and 10 peaches. Then Ilonka picks one apple and one peach. When does Ilonka have more choices? - If Péter picks an apple: apples left = 11, peaches = 10. - Ilonka's choices: $11 \times 10 = 110$. - If Péter picks a peach: apples = 12, peaches = 9. - Ilonka's choices: $12 \times 9 = 108$. - Ilonka has more choices if Péter picks an apple. 4. **Problem:** How many 3-color striped flags can be made from 5 colors including red? How many have red? - Total 3-color flags from 5 colors: permutations $P(5,3) = 5 \times 4 \times 3 = 60$. - Flags without red: $P(4,3) = 4 \times 3 \times 2 = 24$. - Flags with red: $60 - 24 = 36$. 5. **Problem:** How many dictionaries are needed to translate directly between Russian, English, French, German, Italian? - Number of languages: 5. - Number of language pairs (ordered): $5 \times (5-1) = 20$. 6. **Problem:** How many dictionaries needed if languages are 10? - Number of pairs: $10 \times 9 = 90$. 7. **Problem:** How many ways to select 4 cards from a 52-card deck with all different suits? - There are 4 suits, 13 cards each. - Choose one card from each suit: $13^4 = 28561$. 8. **Problem:** How many ways to select 4 cards with different suits and no two cards have the same rank? - Choose 4 different ranks from 13: $\binom{13}{4} = 715$. - Assign each rank to a different suit: $4! = 24$. - Total ways: $715 \times 24 = 17160$. 9. **Problem:** How many ways to select 4 cards with different suits so that black suits and red suits have the same rank? - Choose 2 ranks for black suits (clubs and spades): $\binom{13}{2} = 78$. - Choose 2 ranks for red suits (hearts and diamonds) same as black: must be the same ranks. - Assign ranks to black suits: $2! = 2$ ways. - Assign ranks to red suits: $2! = 2$ ways. - Total ways: $78 \times 2 \times 2 = 312$. 10. **Problem:** How many name sequences can a child have if there are 300 names and at most 3 names per person? - Number of sequences with 1 name: $300$. - With 2 names: $300^2 = 90000$. - With 3 names: $300^3 = 27000000$. - Total: $300 + 90000 + 27000000 = 27090300$. 11. **Problem:** How many ways to seat 4 people around a round table, considering rotations the same? - Number of seatings: $(4-1)! = 6$. 12. **Problem:** How many seatings of 7 people with 2 designated people sitting together? - Treat the 2 as one unit: $6! = 720$. - The 2 can swap seats: $2$ ways. - Total: $720 \times 2 = 1440$. 13. **Problem:** 5 girls and 3 boys form two teams of 4, each with at least one boy. - Total ways to split into two 4-person teams: $\frac{\binom{8}{4}}{2} = 35$. - Count teams with no boys and subtract. - Teams with no boys: only girls, impossible since 5 girls and 3 boys. - Calculate number of teams with at least one boy in each team: 30. 14. **Problem:** Distribute 6 letters among 3 postmen. - Each letter can go to any of 3 postmen: $3^6 = 729$ ways. 15. **Problem:** 5 people A,B,C,D,E speak in order, B cannot speak before A. - Total permutations: $5! = 120$. - Half have B before A, half have A before B. - Valid permutations: $120/2 = 60$. 16. **Problem:** Seat 5 men and 5 women around a round table so no two men or two women sit together, rotations considered different. - Arrange men in $5! = 120$ ways. - Place women in gaps: $5! = 120$ ways. - Total: $120 \times 120 = 14400$. 17. **Problem:** Draw 10 cards from a deck. How many have at least one ace? Exactly one ace? - Total ways: $\binom{52}{10}$. - Without aces: $\binom{48}{10}$. - At least one ace: $\binom{52}{10} - \binom{48}{10}$. - Exactly one ace: $\binom{4}{1} \times \binom{48}{9}$. 18. **Problem:** At least two aces? Exactly two aces? - Exactly two aces: $\binom{4}{2} \times \binom{48}{8}$. - At least two aces: $\binom{52}{10} - \binom{48}{10} - \binom{4}{1} \times \binom{48}{9}$. 19. **Problem:** Max population if no two people have the same missing teeth pattern, max 32 teeth. - Each tooth can be present or missing: $2^{32}$ patterns. - Max population: $2^{32} = 4294967296$. 20. **Problem:** Number plates with 1-3 letters (32 letters) and 4-digit numbers. - Letters: $32^1 + 32^2 + 32^3 = 32 + 1024 + 32768 = 33824$. - Numbers: $10^4 = 10000$. - Total plates: $33824 \times 10000 = 338240000$. 21. **Problem:** 2 apples and 3 pears given over 5 days, one fruit per day. - Number of sequences: $\frac{5!}{2!3!} = 10$. 22. **Problem:** 2 apples, 3 pears, 4 peaches over 9 days. - Number of sequences: $\frac{9!}{2!3!4!} = 1260$. 23. **Problem:** Distribute 5 distinct oranges to 8 sons, each gets at most one. - Choose 5 sons: $\binom{8}{5} = 56$. - Assign oranges: $5! = 120$. - Total ways: $56 \times 120 = 6720$. 24. **Problem:** Number of ways to arrange letters of "matematika". - Letters: m(2), a(3), t(2), e(1), i(1), k(1). - Total letters: 10. - Number of arrangements: $\frac{10!}{2!3!2!} = 151200$. 25. **Problem:** Arrange white pieces (2 rooks, 2 knights, 2 bishops, queen, king) on first row. - Total pieces: 8. - Number of arrangements: $\frac{8!}{2!2!2!} = 10080$. 26. **Problem:** Buy 12 postcards from 10 types. - Number of combinations with repetition: $\binom{10+12-1}{12} = \binom{21}{12} = 293930$. 27. **Problem:** Choose 6 from 7 men and 4 women with at least 2 women. - Total ways: $\binom{11}{6} = 462$. - Ways with 0 women: $\binom{7}{6} = 7$. - Ways with 1 woman: $\binom{4}{1} \times \binom{7}{5} = 4 \times 21 = 84$. - Valid ways: $462 - 7 - 84 = 371$. 28. **Problem:** 4-digit numbers divisible by 4 from digits {1,2,3,4,5}, digits can repeat. - Divisible by 4 if last two digits divisible by 4. - Possible last two digits divisible by 4 from digits: 12, 24, 32, 44, 52. - Number of choices for first two digits: $5^2 = 25$. - Number of valid last two digits: 5. - Total numbers: $25 \times 5 = 125$.