1. Problem: Determine if 7 different colors are adequate to generate 42 unique color codes each consisting of 3 colors with no repetition. Step 1: Calculate the number of 3-color combinations from 7 colors, order not important, no repetition. This is a combination problem: $$\binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35.$$ Step 2: Since 35 codes are fewer than 42 parts needed, 7 colors are not adequate to generate 42 unique 3-color codes. 2. Problem: Find the number of different combinations for choosing 4 subjects from 7 subjects. Step 1: This is choosing 4 out of 7 without order, so use combinations: $$\binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35.$$ Step 2: There are 35 different ways to choose 4 subjects from 7. 3. Problem: Find how many different committees of 5 members can be formed from 15 members. Step 1: This is choosing 5 out of 15, combination without order: $$\binom{15}{5} = \frac{15!}{5!(15-5)!} = \frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} = 3003.$$ Step 2: There are 3,003 possible committees. Final answers: 1. Not adequate, only 35 codes can be generated. 2. 35 combinations. 3. 3,003 committees.