#1 Counting Principles

There are \(6+5=11\) cards in total, so one card can be chosen in \(11\) ways.

For each of the \(4\) soups, there are \(3\) choices of main dish. Total: \(4\cdot3=12\) lunches.

The tens digit has \(5\) choices, and the units digit has \(4\) choices. Total: \(5\cdot4=20\).

Each element is either included or not, so there are \(2^6=64\) subsets. One is empty, so \(64-1=63\) are nonempty.

The number \(a\) can be \(1,2,3,4,5,6\). After choosing \(a\), \(b=7-a\) is positive and determined. Answer: \(6\).

If the last digit is \(0\), the hundreds digit has \(5\) choices and the tens digit \(4\): \(20\). If the last digit is \(2\) or \(4\), the hundreds digit cannot be \(0\) or the last digit: \(4\) choices, then \(4\) for the tens digit. For two such cases: \(2\cdot4\cdot4=32\). Total: \(52\).

We must place \(2\) up steps among \(5\) total steps. This can be done in \(10\) ways: choose the two positions of the up steps among five.

First letter: \(3\) choices. Each of the next three letters: \(2\) choices. Total: \(3\cdot2^3=24\).

Multiples of \(3\): \(66\). Multiples of \(5\): \(40\). Multiples of both, i.e. \(15\): \(13\). Answer: \(66+40-13=93\).

The chair can be chosen in \(12\) ways, then the secretary in \(11\) ways. Total: \(12\cdot11=132\).

There are \(35\) teams of \(3\) from \(7\) people. All-boy teams: \(4\). Therefore the answer is \(35-4=31\).

Choose the first stripe in \(4\) ways, the second in \(3\), and the third in \(3\). Total: \(4\cdot3\cdot3=36\).

If the last digit is \(0\), the first has \(7\) choices, then \(6\) and \(5\): \(210\). If the last digit is \(5\), the first is chosen from \(1,2,3,4,6,7\): \(6\) choices, then \(6\) and \(5\): \(180\). Total: \(390\).

Choose \(3\) positions out of \(10\) for the ones; all remaining positions are zeros. This gives \(120\) strings.

Total: \(3^5=243\). Without \(A\): \(2^5=32\). Without \(B\): \(32\). Without both letters: only \(CCCCC\), one word. Answer: \(243-32-32+1=180\).

There are \(35\) paths in total: choose the positions of three up moves among seven. Through \((2,1)\): \(3\) paths to the point and \(6\) paths from it to \((4,3)\). Forbidden paths: \(18\). Answer: \(35-18=17\).

Among \(1\) to \(20\), there are \(10\) even and \(10\) odd numbers. Pairs with even sum: both even \(10\cdot10\) or both odd \(10\cdot10\). Total: \(200\).

Choose any \(3\) digits from \(1,2,\ldots,9\). After choosing them, there is exactly one increasing order. There are \(84\) such choices.

After placing one ball in each box, \(2\) balls remain. Possibilities: both in one box — \(3\) ways; one in each of two boxes — \(3\) ways. Total: \(6\).

A \(4\) by \(5\) grid has \(5\) horizontal and \(6\) vertical grid lines. Choose \(2\) vertical and \(2\) horizontal lines: \(15\cdot10=150\).

There are \(5\) even and \(5\) odd decimal digits. For six digits, the digit sum is even if the number of odd digits is \(2\) or \(4\). With \(2\) odd digits, there are \(10\) sets without zero and \(40\) with zero; contribution \(10\cdot720+40\cdot600=31200\). With \(4\) odd digits, there are \(30\) sets without zero and \(20\) with zero; contribution \(30\cdot720+20\cdot600=33600\). Total: \(64800\).

There are \(2^{20}\) subsets. With no even number: choose only from \(10\) odd numbers, \(2^{10}\). With no multiple of \(5\): \(16\) available numbers, \(2^{16}\). With neither even numbers nor multiples of \(5\): \(8\) odd nonmultiples of \(5\), so \(2^8\). Answer: \(2^{20}-2^{10}-2^{16}+2^8\).

If all \(5\) symbols are equal, there are \(4\) strings. If the multiplicities are \(3\) and \(2\), choose the letter appearing \(3\) times: \(4\) ways, the letter appearing \(2\) times: \(3\) ways, and the positions of the first letter: \(10\) ways. Total: \(4+4\cdot3\cdot10=124\).

Consider the \(7\) possible forbidden adjacencies. If \(k\) adjacencies are chosen, they form \(c\) blocks of consecutive edges in a path. Such \(k\) edges can be chosen in \(C(k-1,c-1)C(8-k,c)\) ways. Each block has \(2\) orientations, so \(2^c\) orientations, and after gluing there are \(8-k\) objects to permute, giving \((8-k)!\). Inclusion-exclusion gives \(40320-70560+51840-20400+4608-612+48-2=5242\).