Permutations

Treat the vowels as one block, giving (5!) arrangements and (3!/2!) internal arrangements. In exams, always check repeated letters.

There are (6) letters with (S) repeated four times, so the number is (6!/4!). In exams, do not forget to divide by repeated letters.

The last digit must be (0) or (5), and the leading zero case must be counted separately. For such questions, start with the unit digit.

Subtract the together cases (2\cdot 6!) from the total (7!). For restriction questions, subtracting bad cases is often easier.

Treat the two people as one block, so (7) units have circular arrangements (6!), with (2!) internal ways. In circular arrangements, rotations are identical.

There are (4) possible position pairs for (T) and (E), with (2) orders, and the remaining (6!) arrangements. In position-based questions, choose fixed positions first.

The last digit can be (2,4,6) in (3) ways, and the remaining four places in (^{6}P_{4}) ways. For even numbers, fix the unit digit first.

There are (9) letters with (M,T,E) each repeated twice, so the count is (9!/(2!2!2!)). For difficult words, list letter counts first.

Arrange the boys in (5!) ways, then place the girls in (^{6}P_{4}) ways in (4) of the (6) gaps. The gap method is safest for such questions.

The two specified books occupy the ends in (2!) ways, and the remaining (4) books in (4!) ways. Fill the most restricted places first.

In the total (6!) arrangements, the two orders of (A) and (B) are equally likely, so half are valid. Symmetry saves time in order-condition questions.

There are (4) odd positions among (8), and the (4) vowels occupy them in (4!) ways, while consonants also arrange in (4!) ways. When counts match, arrange groups separately.

The digit sum must be divisible by (3); arrange valid (5)-digit sets and subtract leading-zero arrangements. Choosing digit sets first is useful.

For a necklace, the count is ((9-1)!/2) because both rotation and reflection are identical. Remember the difference between circular seating and necklaces.

Add arrangements beginning with smaller available letters at each position, then add (1). In rank questions, write the letters in alphabetical order first.

This is the derangement of (4) objects, whose value is (9). For such questions, use the derangement formula or a small listing.

This is ordered selection, so (^{10}P_{3}=10\cdot 9\cdot 8). When positions are ranked, order matters.

The three offices are different, so assigning the same three people in different offices creates different outcomes. This is an ordered selection, not an ordinary combination. Choose the president in 8 ways, then the secretary from the remaining 7 people, and finally the treasurer from the remaining 6 people. The condition that nobody holds two offices is automatically respected by reducing the choices after each selection.

The total number is \(8\times7\times6=336\), equivalently \({}^{8}P_{3}=\frac{8!}{(8-3)!}\). Since president, secretary, and treasurer are distinct roles, order matters. Therefore option A, 336, is correct. The value 56 would be \(\binom{8}{3}\), which chooses people without assigning offices and is therefore insufficient.

There are (7) letters with (L) and (O) each repeated twice, so the count is (7!/(2!2!)). Repeated letters are not treated as distinct objects.

Treat the two (L)'s as one block, giving (6) units with (O) repeated twice, so the count is (6!/2!). Use the block method for together conditions.

The thousands digit can be (4,5,6,7,8), and the remaining (3) places are filled in (^{7}P_{3}) ways. In comparison questions, fix the first digit first.

There are two possible patterns, and in each pattern men arrange in (6!) ways and women in (6!) ways. For equal alternate seating, remember the two patterns.

The total distinct arrangements are (7!/2!), and among the possible vowel orders only one valid alphabetical order is allowed. Count order restrictions carefully when vowels repeat.

This asks for the number of orders in which 5 distinct keys can be tried. There are 5 choices for the first position, then 4, 3, 2, and 1 choices. Thus, the total number of orders is \(5\times4\times3\times2\times1=5!=120\). Option 60, or \(5!/2\), would arise only if an extra condition made two orders equivalent; no such condition is given here. Exam tip: use \(n!\) when all \(n\) distinct objects are arranged once in order.

Subtract the together cases (5!) from the total (6!/2!). For not-together conditions, subtract together arrangements from total.