There are 14 chairs and 8 people to be seated. But among the 8. three will be seated together:
So 5 people and (3) could be considered as 6 entities:
Since the order matters, we have to use permutation:
¹⁴P₆ = (14!)/(14-6)! = 2,162,160, But the family composed of 3 people can permute among them in 3! ways or 6 ways. So the total number of permutation will be ¹⁴P₆ x 3!
2,162,160 x 6 = 12,972,960 ways.
Another way to solve this problem is as follow:
5 + (3) people are considered (for the time being) as 6 entities:
The 1st has a choice among 14 ways
The 2nd has a choice among 13 ways
The 3rd has a choice among 12 ways
The 4th has a choice among 11 ways
The 5th has a choice among 10 ways
The 6th has a choice among 9ways
So far there are 14x13x12x11x10x9 = 2,162,160 ways
But the 3 (that formed one group) could seat among themselves in 3!
or 6 ways:
Total number of permutation = 2,162,160 x 6 = 12,972,960
Answer:
1. 12
2. 4
3. 6.8
4. 4.5
5 1.2
Step-by-step explanation:
Answer:
they each share the same variable which is raised to the same exponent, 1
Step-by-step explanation:
Answer:
The probability that a student chosen at random is fluent in English or Swahili.
P(S∪E) = 1.1
Step-by-step explanation:
<u><em>Step(i):</em></u>-
Given total number of students n(T) = 150
Given 125 of them are fluent in Swahili
Let 'S' be the event of fluent in Swahili language
n(S) = 125
The probability that the fluent in Swahili language
Let 'E' be the event of fluent in English language
n(E) = 135
The probability that the fluent in English language
n(E∩S) = 95
The probability that the fluent in English and Swahili
<u><em>Step(ii):</em></u>-
The probability that a student chosen at random is fluent in English or Swahili.
P(S∪E) = P(S) + P(E) - P(S∩E)
= 0.833+0.9-0.633
= 1.1
<u><em>Final answer:-</em></u>
The probability that a student chosen at random is fluent in English or Swahili.
P(S∪E) = 1.1
I have attached the correct answer. Brainliest will be much appreciated!
