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i think the last one we need to find 2 different colours which will be denoted by 4!/2!2! and as we need 2 for each colour so the 6 different colours is to be selected for 2 holes so 6c2 so the answer is 90 and then add all of the previous answers to it
Explain this please (7*6*5*(6!/3!)).
Oh! JazakAllah brother
What is the answer?
But then you are missing when 3Gs are together.
No i mean what is the actual answer to this question, i got 5040
Remember the fi thing i said before, use that formula
Yea it is 5040. I got the same when I did 7*6*(6!/3!). What I wanna do is total no of arrangements - arrangements when none of the G's are next to each other.No i mean what is the actual answer to this question, i got 5040
Im not sure. but i think u can take probability at 63 to be 0.75 since it is upper quartile. u do know the mean.
I know I won't get 5040 but why is it negative? Why is the number of arrangements when none of the G's are next to each other more than the total? Where's the mistake?
there also have to arrangements when 3 Gs are together then.
THANXX A MIllion tyms.........solved it 10 times........bt neva complete;y undrstoood thats y did mistake every time......absolutely loved that words............"To ensure Mrs Lin behind a student, group her and one student together as a unit. (Note that she cannot switch the seats with that student since she must be behind the student.) For this "unit", there are 10 pairs of seats valid. 10P1"The (i) part is always the most direct in all P & C questions
14P12 = 4.36E^10
(ii) -The 3 seats on the right are taken by the 3 businessmen. So first it's 3P3.
-Then the Lins sit in a row on the same side of the aisle, the available locations could only be the 3 pairs of seats on the upper part. (Front pair excluded since taken byy the businessmen.) Moreover the couple can switch seats with each other. It's 3P1 × 2P2.
-Next are the Browns. 1 pair in the 3 taken by the Lins, so for them, 2P1 × 2P2.
-Finally, the students. You can see there are just the right number of seats left: 5 window seats for 5 students. 5P5.
Answer: 3P3 × 3P1 × 2P2 × 2P1 × 2P2 × 5P5 = 17280
(iii) If they seat randomly, we are back to (i) then. Total number of arrangements is 4.36E^10
-If to ensure that Mrs Brown in the front, there are only 3 seats for her to sit in. 3P1
-To ensure Mrs Lin behind a student, group her and one student together as a unit. (Note that she cannot switch the seats with that student since she must be behind the student.) For this "unit", there are 10 pairs of seats valid. 10P1.
--There a 5 students that can be grouped with Mrs Lin, so for the case above, it's 5 × 10P1.
-Finally the others. There are 14 - 1 -2 = 11 seats for 12 - 1 - 2 = 9 people. 11P9.
Number of arrangements = 3P1 × 5 × 10P1 × 11P9 =2.99E^9
Probability = 2.99E^9 / 4.36E^10 = 25/364 OR 0.0687
I'm not sure if my answers are correct.
You forget one thing, No. of arrangements - No. of arrangements when no G's are next to each other - No. of arrangements when all the G's are together
Do the usual standardizing. just replace µ with 7/3 *σ^2
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