Final answer:
(a) The number of distinct arrangements: ( {14!} / {4! . 3! . 7!} )
(b) Probability of the specific arrangement: {1} / {{14!} / {4! . 3! . 7!}} )
(c) Probability of grouped by color: ( {1} / {{14!} / {4! . 3! . 7!}}
Step-by-step explanation:
(a) To find the number of distinct arrangements, we use the formula for permutations. The total number of cars is 14, with repetitions of colors. The formula for permutations with repetitions is given by:
P(n; n_1, n_2,ldots, n_k) = {n!} / {n_1! . n_2! . ldots . n_k!}
Where:
- n is the total number of items (cars in this case),
- n_1, n_2, ldots, n_k are the repetitions of each type (white, black, blue).
In this case:
P(14; 4, 3, 7) = 14! / 4! . 3! . 7!
Calculate this to find the total number of distinct arrangements.
(b) For the probability of a specific arrangement, we use the same formula but considering only one specific arrangement out of all possible arrangements. So, for the given arrangement of 4 white, 3 black, and 7 blue cars:
Probability = 1 / P(14; 4, 3, 7)
(c) To find the probability that the cars are grouped by color, we consider all possible groupings. There are 3 colors, so the number of arrangements is the product of the permutations for each color:
P(14; 4, 3, 7) = P(4; 4) . P(3; 3) . P(7; 7)
The probability is then:
Probability = 1 / P(14; 4, 3, 7)
Now, let's calculate these values:
P(14; 4, 3, 7) = 14! / 4! . 3! . 7!
P(4; 4) . P(3; 3) . P(7; 7) = {4!} / {4!} . {3!} / {3!} . {7!} / {7!} \]
Finally, calculate the probabilities:
(b) Probability of the specific arrangement: 1 / P(14; 4, 3, 7)
(c) Probability of grouped by color: 1 / P(14; 4, 3, 7)
your complete question is: Fourteen cars of the same style, of which 4,3, and 7, are white black and blue respectively are randomly parked in a row. (a) How many distinct arrangements are possible? (b) what is the probability that a random parking of the cars will result in the first 4 cars being white, the next 3 cars being black, and the last 7 cars being blue? (c) what is the probability that a random parking of cars results in the cars being grouped by color?