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There is a committee to be selected comprising of 5 people from a group of 5 men and 6 women. Whether the selection is randomly done then determines the possibility of having the given possibilities combinations
i. One particular man and one particular woman should not be in the committee (one man four women)
Solution
i. P(one particular man and one particular woman should not be in the committee would be determined as given below:
The group size = 5m + 6w
Committee size = 5 people
Actual groups size from which to
Select the committee = 4m + 5w
Committee = 1m + 4w
The committee may be selected in 9C5
The one man may be selected in 4C1 ways
The four women may be selected in 5C4 ways
∴ P(committee of 4w1man).
(5C4 * 4C1) = 9C5
= 10/63
If tanA+sinA=m and tanA-sinA=n, show that m 2 -n 2 = 4√mn Ans: TanA + SinA = m TanA - SinA = n. m 2 -n 2 =4√mn . m 2 -n 2 = (TanA + SinA) 2 -(TanA - SinA) 2
40.783-75
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