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Answer :
1. The number of different ways to prepare both vegetable and fruit salad is [tex]{450}\)[/tex]. 2. The number of different ways to either a vegetable or fruit salad but not both is [tex]{55}\)[/tex]. 3. The number of ways to choose two different kinds of salad is [tex]{3475}\)[/tex].
1.) The number of combinations for each can be calculated using the combination formula [tex]\( C(n, k) = \frac{n!}{k!(n-k)!} \)[/tex], where n is the total number of items to choose from, k is the number of items to choose and denotes factorial.
For the vegetable salad: [tex]\[ C(5, 3) = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10 \][/tex]
For the fruit salad: [tex]\[ C(10, 2) = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45 \][/tex]
Therefore, the total number of ways to prepare both vegetable and fruit salad is: [tex]\[ 10 \times 45 = 450 \][/tex]
2. To prepare a vegetable salad or a fruit salad but not both, we add the number of ways to prepare just a vegetable salad to the number of ways to prepare just a fruit salad.
For just a vegetable salad: [tex]\[ C(5, 3) = 10 \][/tex]
For just a fruit salad: [tex]\[ C(10, 2) = 45 \][/tex]
Thus, the total number of ways to prepare either a vegetable or fruit salad but not both is [tex]\[ 10 + 45 = 55 \][/tex]
3. Now we need to calculate the combinations for two vegetable salads and two fruit salads.
For two vegetable salads (since we already have the combination for one vegetable salad, we square it): [tex]C(5, 3)^2 = 10^2 = 100[/tex]
For two fruit salads (since we already have the combination for one fruit salad, we square it): [tex]\[ C(10, 2)^2 = 45^2 = 2025 \][/tex]
For one vegetable salad and one fruit salad (which we have already calculated): [tex]\[ C(5, 3) \times C(10, 2) = 10 \times 45 = 450 \][/tex]
Adding these up gives us the total number of ways to choose two different kinds of salad: [tex]\[ 100 + 2025 + 450 = 3475 \][/tex]
Therefore, the final answers are:
1. The number of different ways to prepare both vegetable and fruit salad is 450.
2. The number of different ways either vegetable or fruit salad but not both is 55.
3. The number of ways to choose two different kinds of salad is 3475.
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Answer:
1). 450
2). 55
3). 1485
Step-by-step explanation:
Out of 5 different vegetables number of ways to select 3 different vegetables = [tex]^{5}C_{3}=\frac{5!}{3!\times (5-3)!}[/tex]
= [tex]\frac{5!}{3!\times 2!}[/tex]
= 10
Out of 10 different fruits number of ways to select 2 fruits
= [tex]^{10}C_{2}[/tex]
= [tex]\frac{10!}{2!\times (10-2)!}[/tex]
= [tex]\frac{10!}{2!\times 8!}[/tex]
= [tex]\frac{90}{2}[/tex]
= 45
(1). Different ways to prepare a vegetable AND a fruit salad
= 10×45
= 450
(2). Different ways to prepare a veg salad OR a fruit salad
= 10 + 45
= 55
(3). Number of ways to choose two different kinds of salad
= [tex]^{55}C_{2}[/tex]
= [tex]\frac{55!}{53!2!}[/tex]
= [tex]\frac{55\times 54}{2}[/tex]
= 1485
