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QuestionFind the number of ways to choose 2 distinct items from 10 food items. Options: a. 5 b. 20 c. 25 d. 45

Studdy Solution

STEP 1

Assumptions1. The cafeteria has10 distinct food items. . The student is choosing distinct food items.
3. The order in which the food items are chosen does not matter.

STEP 2

The problem is asking for combinations, not permutations, so we will use the formula for combinations, which isC(n,k)=n!k!(n−k)!C(n, k) = \frac{n!}{k!(n-k)!}where- n is the total number of items, - k is the number of items to choose, - ! denotes factorial, which is the product of all positive integers up to that number.

STEP 3

Now, plug in the given values for n and k into the formula.
C(10,2)=10!2!(10−2)!C(10,2) = \frac{10!}{2!(10-2)!}

STEP 4

implify the equation.
C(10,2)=10!2!8!C(10,2) = \frac{10!}{2!8!}

STEP 5

Calculate the factorial values.
10!=10×9×8×7××5×4×3×2×110! =10 \times9 \times8 \times7 \times \times5 \times4 \times3 \times2 \times12!=2×12! =2 \times18!=8×7××5×4×3×2×18! =8 \times7 \times \times5 \times4 \times3 \times2 \times1

STEP 6

Substitute the factorial values back into the equation.
C(10,2)=10×9×8××6×5×4×3×2×12×1×8××6×5×4×3×2×1C(10,2) = \frac{10 \times9 \times8 \times \times6 \times5 \times4 \times3 \times2 \times1}{2 \times1 \times8 \times \times6 \times5 \times4 \times3 \times2 \times1}

STEP 7

implify the equation by cancelling out the common factors in the numerator and the denominator.
C(10,2)=10×92×1C(10,2) = \frac{10 \times9}{2 \times1}

STEP 8

Calculate the final value.
C(10,2)=902=45C(10,2) = \frac{90}{2} =45So, a student can choose45 combinations of two distinct food items from a cafeteria with ten food items.

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