Permutations

Arrange the boys around the circle in ((4-1)!) ways and place the girls in the (4) gaps in (4!) ways. For circular alternate seating, fix one group first.

Here (E,T,I) are each repeated twice, so the count is (10!/(2!2!2!)). Divide by the factorial of each repeated letter.

Inside the vowel block there are (5!/(2!2!)) arrangements, and outside there are (6!/2!) arrangements. Do not forget repeated letters when forming blocks.

The four odd digits occupy the four odd positions in (4!) ways, and even digits fill the rest in (4!) ways. Identify the type of positions first.

For a bracelet, the count is ((10-1)!/2) because both rotation and reflection are identical. Remember the difference between bracelets and circular arrangements.

The last digit is (0) or (5), and if (5) is last, (0) must be included inside. Count divisibility and compulsory digit cases separately.

Among the (6) relative orders of (A,B,C), (A) is first in (2) orders, so the valid fraction is (4/6). Relative order solves such questions quickly.

Arrange the men in (6!) ways and place the (4) women in (^{7}P_4) ways in the (7) gaps. Use the gap method for non-adjacent conditions.

In increasing order, each chosen set gives only one arrangement, and including (0) would not give a (6)-digit number. So choose (6) digits from (1) to (9).

Add arrangements starting with smaller available letters at each position, then add (1). In rank problems, first arrange letters alphabetically.

This is the derangement of (5) objects, whose value is (44). In derangement, no object remains in its correct place.

Choose the (2) wrong letters in (^{6}C_2) ways, and they must swap with each other. For exactly (2) wrong, those two letters interchange.

Here (A) is repeated twice and (L) three times, so the count is (8!/(2!3!)). Count repeated letters first.

Treat the three (L)'s as one block, giving (6) units with (A) repeated twice, so the count is (6!/2!). A block of identical letters acts as one unit.

There are (2) ends for (C), and for each, (8) ordered position pairs for (A,B), followed by (5!) arrangements. Fix the end and distance first.

Treat the two specified letters as one block, choose (5) more letters from the remaining (8), and arrange (6) units. Include the (2!) internal orders of the block.

The three odd positions can be filled from (4) odd digits in (^{4}P_3) ways, and the remaining three positions from (5) remaining digits in (^{5}P_3) ways. Apply position restrictions first.

The number of arrangements is (12!/(4!3!2!)). Divide by the factorial of each group of identical objects.

Total circular arrangements are (8!), and adjacent cases are (2\cdot 7!), so the difference is (30240). In circular arrangements, start with ((n-1)!).

It has (N) three times, (E) three times and (D) twice, so the count is (11!/(3!3!2!)). Write frequencies before applying the formula.

Place the (6) white balls first, creating (7) gaps, and choose (4) of them for black balls. With identical objects, only position selection is needed.

Choose the correctly assigned person in (8) ways, and derange the remaining (7) people with (D_7=1854). Exactly one correct means all others are wrong.

Choose the even and odd digit groups and count (7!) arrangements, then subtract leading-zero cases. When (0) is included, handle the first digit carefully.

Arrange consonants in (7!/(2!2!)) ways, then choose (4) of the (8) gaps and arrange vowels in (4!/(2!2!)) ways. Use the gap method for non-adjacent vowels.

Any chosen set of (5) digits has only one decreasing arrangement, and the first digit will not be zero. So choose (5) digits from (10).