A student has 10 blue pens, 5 red pens, and 7 black pens. In how many ways can he select 5 pens if only the choice of colors matters. Use the Counting Rules.
Q. A student has 10 blue pens, 5 red pens, and 7 black pens. In how many ways can he select 5 pens if only the choice of colors matters. Use the Counting Rules.
Initial Setup: Since the choice of colors is the only thing that matters, we can ignore the number of pens of each color. The student can choose from blue, red, or black pens. We need to find the combinations of 5 pens from these 3 colors.
Color Choices: There are 3 choices for the first pen, 3 choices for the second pen, and so on, until the fifth pen. Since the order doesn't matter, we use the combination formula which is C(n,k)=k!(n−k)!n!, where n is the total number of items, and k is the number of items to choose.
Combination Formula: However, since we are not distinguishing between pens of the same color, we actually have a simpler problem. We just need to count the number of ways to distribute 5 identical items into 3 distinct groups. This is a problem of combinations with repetition, also known as stars and bars.
Combinations with Repetition: The formula for combinations with repetition is C(n+k−1,k), where n is the number of types of items (colors of pens in this case) and k is the number of items to choose (pens). Here, n=3 and k=5.
Plug Values: Plug the values into the formula: C(3+5−1,5)=C(7,5). Now we calculate C(7,5) using the combination formula: C(7,5)=(5!(7−5)!)7!=(5!2!)7!.
Calculate Factorials: Calculate the factorial values: 7!=7×6×5!, 5!=5×4×3×2×1, and 2!=2×1. So, C(7,5)=5!×2×17×6×5!.
Simplify Expression: Simplify the expression by canceling out the common 5! terms: C(7,5)=2×17×6=242.
Final Calculation: Finish the calculation: 42/2=21. So, there are 21 different ways to select 5 pens when only the choice of colors matters.
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