Permutations

Correct answer: C. (240)

Treat vowels (A,I) as one block, giving (5!\cdot 2!) arrangements. For together conditions, the block method is most useful.

Correct answer: D. (300)

The total arrangements are (7!/(3!2!)), and vowel-together cases are (120). For not-together cases, subtract together cases from total.

Correct answer: B. (3780)

The first digit must be one of (5,6,7,8), and the last digit must be even, so count by cases. In such problems, fix the first and last positions first.

Correct answer: A. (4320)

Treat the three people as one block, so (7) units have (6!) circular arrangements and (3!) internal ways. In circular arrangements, rotations are identical.

Correct answer: D. (3360)

Here (E) is repeated three times and (N) twice, so the count is (8!/(3!2!)). Always divide by repeated letters.

Correct answer: B. (240)

The direct answer is B: 240. ENGINEER has four vowels: E, E, E and I, and four consonants: N, N, G and R. To keep all vowels together, treat the four vowels as one block. Then there are five objects: the vowel block, N, N, G and R. These can be arranged in \(5!/2!=60\) ways because N is repeated. Inside the vowel block, the arrangements are \(4!/3!=4\), because E is repeated three times. Therefore the total is \((5!/2!)(4!/3!)=60\times4=240\). Option B is correct. Option A, 180, omits valid orders. Option C, 360, overcounts one repeated group. Option D, 480, overcounts further. The block method must be combined with repetition corrections.

Correct answer: C. (36000)

Subtract the cases where all three are together, (6!\cdot 3!), from the total (8!). For not-all-together conditions, subtract bad cases from total.

Correct answer: A. (1260)

In the total (7!) arrangements, divide by (4) because of two independent order conditions. Symmetry gives a quick answer in such problems.

Correct answer: D. (129600)

Letters can be arranged in (^{6}P_4) ways, and digits with an even last digit in (5\cdot 9\cdot 8) ways. Count different parts of a code separately.

Correct answer: B. (604800)

Among the (3!) relative orders of (A,B,C), only one is valid, so the count is (10!/3!). For relative order, divide by the number of possible orders.

Correct answer: C. (35)

Add arrangements formed by smaller available letters at each position, then add (1). In rank problems, handle repeated letters carefully.

Correct answer: A. (28800)

There are two possible patterns, and in each pattern boys arrange in (5!) ways and girls in (5!) ways. With equal numbers, alternate seating has two starts.

Correct answer: D. (7680)

Treat each couple as one block, so (6) blocks have (5!) circular arrangements and each block has (2) internal ways. For couple problems, remember the (2^n) internal orders.

Correct answer: B. (1680)

In the total (8!) arrangements, the (4!) relative orders of consonants are equally likely, so the count is (8!/4!). For original-order conditions, divide by relative orders.

Correct answer: C. (924)

The last two digits can be (25,50,75), and leading-zero cases must be removed separately. For divisibility by (25), check the last two digits first.

Correct answer: A. (34650)

It has (I) four times, (S) four times and (P) twice, so the count is (11!/(4!4!2!)). For long words, make a frequency table.

Correct answer: D. (6300)

Treat both (P)'s as one block, giving (10) units with (I) and (S) each repeated four times. Hence the count is (10!/(4!4!)).

Correct answer: B. (14400)

The positions of (A,B) must differ by at least (4), giving (10) position pairs and (2) orders. The remaining (6!) people are arranged afterward.

Correct answer: C. (10920)

The first digit must be (1,2,3), and the last digit must be odd, so cases are needed. Count the bound and oddness together.

Correct answer: A. (720)

Fix (A), then the opposite place for (B) is fixed, and the remaining (6) people arrange in (6!) ways. For opposite seating, fix one person first.

Correct answer: D. (252000)

Choose letter positions in (^{6}C_2) ways, letters in (^{5}P_2) ways and digits in (^{7}P_4) ways. It is easier to decide position types first.

Correct answer: B. (282240)

Subtract the consecutive cases (2\cdot 8!) from the total (9!). For non-consecutive conditions, subtract the bad block cases.

Correct answer: A. (9979200)

Here (B) and (I) are repeated twice, so the count is (11!/(2!2!)). Identifying repeated letters is the first step.

Correct answer: C. (241920)

Inside the vowel block there are (4!/2!) arrangements, and outside there are (8!/2!) arrangements. Divide for repeated consonants and vowels.

Correct answer: D. (840)

Among the (3!) relative orders of the three specified books, only one is valid, so the count is (7!/3!). For relative-order restrictions, divide directly.