Subjects combinatorics

Tortilla Arrangements Fa423A

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Tortilla Arrangements Fa423A


1. **Problem statement:** We have 2 different meat tortillas and 4 different vegetable tortillas to arrange on a plate. We want to find the number of ways to arrange them under different conditions. 2. **General formula for permutations:** The number of ways to arrange $n$ distinct items is $n!$. --- ### a) No additional restrictions 3. Total tortillas: $2 + 4 = 6$. 4. Number of ways to arrange all 6 distinct tortillas is $$6! = 720.$$ --- ### b) The 2 meat tortillas must be together 5. Treat the 2 meat tortillas as a single block. Then we have this block + 4 vegetable tortillas = 5 items. 6. Number of ways to arrange these 5 items is $$5! = 120.$$ 7. Inside the meat block, the 2 meat tortillas can be arranged in $$2! = 2$$ ways. 8. Total arrangements = $$5! \times 2! = 120 \times 2 = 240.$$ --- ### c) The 4 vegetable tortillas must be together 9. Treat the 4 vegetable tortillas as a single block. Then we have this block + 2 meat tortillas = 3 items. 10. Number of ways to arrange these 3 items is $$3! = 6.$$ 11. Inside the vegetable block, the 4 tortillas can be arranged in $$4! = 24$$ ways. 12. Total arrangements = $$3! \times 4! = 6 \times 24 = 144.$$ --- ### d) The 2 meat tortillas must be at the beginning and end of the row 13. The row has 6 positions: positions 1 and 6 must be meat tortillas. 14. Number of ways to arrange 2 meat tortillas in positions 1 and 6 is $$2! = 2.$$ 15. The 4 vegetable tortillas fill positions 2, 3, 4, 5 and can be arranged in $$4! = 24$$ ways. 16. Total arrangements = $$2! \times 4! = 2 \times 24 = 48.$$ --- ### e) The 2 meat tortillas must be together and the 4 vegetable tortillas must be together 17. Treat the 2 meat tortillas as one block and the 4 vegetable tortillas as another block. Now we have 2 blocks to arrange. 18. Number of ways to arrange these 2 blocks is $$2! = 2.$$ 19. Inside the meat block, 2 tortillas can be arranged in $$2! = 2$$ ways. 20. Inside the vegetable block, 4 tortillas can be arranged in $$4! = 24$$ ways. 21. Total arrangements = $$2! \times 2! \times 4! = 2 \times 2 \times 24 = 96.$$ **Final answers:** - a) 720 - b) 240 - c) 144 - d) 48 - e) 96