Fundamental principle of counting

Fix one particular person; then the other particular person has (2) valid positions. Arrange the remaining (6) persons in (6!) ways.

Place (3) of the (4) odd digits in positions (1,3,5) in order. Then fill the even positions from the remaining digits.

First choose (2) doctors, then choose the remaining (5) members from engineers and teachers with at least (2) engineers. Keep exact and at least conditions in separate stages.

Treat all vowels as one block and account for repeated consonants. Count the arrangements inside the vowel block separately.

Subtract arrangements containing the block (12) from total (^{7}P_6). Only the fixed order (1) immediately before (2) is forbidden.

First count positive solutions and subtract solutions where the eldest child gets (5) or more coins. Complement counting is useful for upper-bound restrictions.

The first position has (3) choices, the last has (4), and each middle position has (7) choices. Multiply choices for independent positions.

Among total (10!) arrangements, the (3!) relative orders of the three particular delegates are equally likely. Hence only one order is counted.

First choose (3) rows and (3) columns, then match them in (3!) ways. Matching is useful for row-column restrictions.

A function from a 4-element set to a 5-element set assigns one target element to each of the four source elements. For each source element there are 5 choices, and the choices are independent, so the total number of functions is \(5^4\). A one-one function uses different target elements for different source elements; its count is the permutation number \({}^{5}P_4\). Subtracting gives the functions that are not one-one.

Calculate the total: \(5^4=625\). Calculate the injective functions: \({}^{5}P_4=5\times4\times3\times2=120\). Hence the number that are not one-one is \(625-120=505\). Therefore option B is correct. The result is not found by subtracting only one or two arrangements; every non-injective function is included in the difference between all functions and one-one functions.

The letters have (7^3) choices. For the sum of three distinct digits to be odd, the number of odd digits must be (1) or (3).

First choose exactly (2) poetry books, then choose the remaining (3) books from novels and dramas with at least (1) novel. After selection, arrange in (5!) ways.

First seat the men around the circle in ((6-1)!) ways and place the (4) women in the (6) gaps. Use the gap method for circular separation.

Take cases for the even units digit because the thousands digit must be from (5) to (9). Decide the leading and last positions together.

For the sum to be odd and the product to be even, exactly (1) odd and (2) even numbers must be chosen. Classify parity conditions first.

There are (5) possible positions for the block of three digits, then arrange digits and letters in order. Treat consecutive objects as a block.

The only possible size distributions are ((2,2,1,1)) and ((2,2,2,0)). Count team choices and project assignments for each pattern.

There are two possible patterns, starting with a boy or starting with a girl. In each pattern, boys and girls are arranged separately in (5!) ways.

First divide (8!) by (3!) for the relative order of (A,B,C), then subtract cases where (D) is at an end. Relative orders are equally distributed.

First consider (6)-digit sets that can be split into two (3)-digit parts with equal sums, then count arrangements inside both parts. For equal-sum conditions, think of sets and partitions first.

First choose the (2) children who sit correctly, then derange the remaining (3) children. Exact fixed-position conditions use derangements.

Place the (4) zeros in the (4) gaps around the (3) ones, with no gap receiving more than (2) zeros. Control forbidden consecutiveness by gap limits.

First form the committee in (\binom{8}{3}\binom{6}{3}) ways, then choose (2) distinct posts from (6) members in order. Committee selection and post selection are separate stages.

The third book has (2) end choices, and the two adjacent books are treated as a block with the remaining objects. Count the (2!) internal orders of the block too.

Choose the other (3) team members from (8), then choose the ordered speakers from those (3) members. Fix the compulsory but ineligible member first.