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There are 180 ways to form such committees. Solution: $\binom{4}{2} \times \binom{6}{3} = 6 \times 20 = 120$

There are 100,000 different combinations. Solution: $10^5 = 100,000$

There are 56 different triangles. Solution: $\binom{8}{3} = 56$

There are 6 ways to select the group of balls. Solution: $\binom{5}{1} \times \binom{3}{1} \times \binom{4}{1} = 5 \times 3 \times 4 = 60$

There are 150 ways to distribute the toys. Solution: $\text{Apply Stirling numbers of the second kind } S(5,3) = 25, \text{then multiply by } 3! \text{ to account for the distinctness of toys} = 25 \times 6 = 150$

There are 128 subsets. Solution: $2^7 = 128$

There are 40 different ways to pick a pair. Solution: $8 \times 5 = 40$

There are 265 ways. This is a derangement problem, and the number can be calculated as $!6 = 265$.

There are 30 different 4-letter words. Solution: $\frac{4!}{2!} = 12$ (accounting for the 2 'E's).

There are 210 such binary strings. Solution: $\binom{10}{4} = 210$

There are 120 different ways to arrange the 5 books. Solution: $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

There are 67,600 such passwords. Solution: $26 \times 26 \times 10 \times 10 = 67,600$

There are 125 3-digit numbers. Solution: $5 \times 5 \times 5 = 125$

A decagon has 35 diagonals. Solution: $\frac{10 \times (10-3)}{2} = 35$

There are 720 ways to award the medals. Solution: $10 \times 9 \times 8 = 720$

There are 120 ways to seat the 6 people. Solution: $(6 - 1)! = 5! = 120$

There are 5040 possible passwords. Solution: $10 \times 9 \times 8 \times 7 = 5040$

There are 60 unique signals. Solution: $\frac{6!}{2! \times 3!} = 60$

There are 5,040 ways to seat the people. Solution: $7! = 5,040$ once you fix one person to remove symmetry.

There are 120 such codes. Solution: $6 \times 5 \times 4 = 120$

There are 210 increasing sequences. Solution: $\binom{10}{4} = 210$

There are 17,280 ways to arrange the books. Solution: Arrange the sections (biographies, novels, dictionaries) $3! = 6$ ways and the books within each section $5! \times 3! \times 2! = 1440$ and then multiply, $6 \times 1440 = 17,280$

There are 35 ways to distribute the apples. Solution: $\binom{3+5-1}{3} = \binom{7}{3} = 35$

There are 56 ways to choose the scoops. Solution: $\binom{6+3-1}{3} = \binom{8}{3} = 56$

There are 112 committees possible. Solution: $\binom{10}{3} - \binom{8}{1} = 120 - 8 = 112$

There are 132 ways to elect a president and vice president. Solution: $12 \times 11 = 132$

There are 6 surjective functions. Solution: $\text{Apply the principle of inclusion-exclusion: } 2^3 - \binom{2}{1}(1^3) = 8 - 2 = 6$

There are 2,598,960 different 5-card hands. Solution: $\binom{52}{5} = 2,598,960$