Permutations with repeats and restrictions

Alike objects. If some objects are identical, swapping them makes no new arrangement, so we must divide out those swaps.

• n objects with p₁ alike of one kind, p₂ alike of another, ...: n!/(p₁! p₂! ...) arrangements.
• BANANA has 6 letters with three A's and two N's: 6!/(3! 2!) = 720/12 = 60.
• The digits 1, 1, 2, 2, 3 make 5!/(2! 2!) = 30 five-digit numbers.

Objects that must stay together: glue them into one block, arrange the block with the rest, then arrange the objects inside the block.

• PENCIL with the vowels E and I together: the block (EI) and P, N, C, L make 5 objects, giving 5! = 120 arrangements.
• Inside the block, E and I can swap: × 2! = 2.
• Total 240.

Never together is easier as 'all arrangements minus together': PENCIL has 6! = 720 arrangements in all, so 720 – 240 = 480 keep E and I apart.

Fixed positions: put the fixed objects in place and arrange only the rest. BANANA words starting with B: the remaining A, A, A, N, N give 5!/(3! 2!) = 10.

Digits with conditions. How many 4-digit numbers greater than 5000 can be made from 0, 3, 5, 7 with no repetition?

• The first digit must be 5 or 7: 2 ways (0 is also not allowed there).
• The other three places take the remaining three digits in 3! = 6 ways.
• Total 2 × 6 = 12.