Answer: A) 2880 Explanation: There are total 9 places out of which 4 are even and rest 5 places are odd.   4 women can be arranged at 4 even places in 4! ways.   and 5 men can be placed in remaining 5 places in 5! ways.   Hence, the required number of permutations  = 4! x 5! = 24 x 120 = 2880

Answer: B) 360 Explanation: There are 7 digits 1, 2, 0, 2, 4, 2, 4 in which 2 occurs 3 times, 4 occurs 2 times.    Number of 7 digit numbers = 7!3!×2! = 420   But out of these 420 numbers, there are some numbers which begin with '0' and they are not 7-digit numbers. The number of such numbers beginning with '0'.   =6!3!×2! = 60   Hence the required number of 7 digits numbers = 420 - 60 = 360

Answer: C) 246 Explanation: A Committee of 5 persons is to be formed from 6 gentlemen and 4 ladies by taking.    (i) 1 lady out of 4 and 4 gentlemen out of 6  (ii) 2 ladies out of 4 and 3 gentlemen out of 6  (iii) 3 ladies out of 4 and 2 gentlemen out of 6  (iv) 4 ladies out of 4 and 1 gentlemen out of 6    In case I the number of ways = C14×C46 = 4 x 15 = 60  In case II the number of ways = C24×C36 = 6 x 20 = 120  In case III the number of ways = C34×C26 = 4 x 15 = 60 In case IV the number of ways = C44×C16 = 1 x 6 = 6    Hence, the required number of ways = 60 + 120 + 60 + 6 = 246

Answer: B) 535 Explanation: The number of points of intersection of 37 lines is C237. But 13 straight lines out of the given 37 straight lines pass through the same point A.   Therefore instead of getting C213 points, we get only one point A. Similarly 11 straight lines out of the given 37 straight lines intersect at point B. Therefore instead of getting C211 points, we get only one point B.    Hence the number of intersection points of the lines is C237-C213-C211 +2 = 535

Answer: C) 3/4 Explanation: Let A, B, C be the respective events of solving the problem and A , B, C be the respective events of not solving the problem. Then A, B, C are independent event ∴A, B, C are independent events Now,  P(A) = 1/2 , P(B) = 1/3 and P(C)=1/4  PA=12, PB=23, PC= 34 ∴ P( none  solves the problem) = P(not A) and (not B) and (not C)                     = PA∩B∩C                    = PAPBPC         ∵ A, B, C are Independent                                          =  12×23×34                     = 14   Hence, P(the problem will be solved) = 1 - P(none solves the problem)                  = 1-14= 3/4

Answer: C) 2/7 Explanation: P( only one of them will be selected) = p[(E and not F) or (F and not E)]   = PE∩F∪F∩E    = PEPF+PFPE    =17×45+15×67=27

Answer: B) 19/42 Explanation: A red ball can be drawn in two mutually exclusive ways  (i) Selecting bag I and then drawing a red ball from it.   (ii) Selecting bag II and then drawing a red ball from it.   Let E1, E2 and A denote the events defined as follows: E1 = selecting bag I, E2 = selecting bag II A = drawing a red ball Since one of the two bags is selected randomly, therefore  P(E1) = 1/2  and  P(E2) = 1/2 Now, PAE1 = Probability of drawing a red ball when the first bag has been selected = 4/7   PAE2  = Probability of drawing a red ball when the second bag has been selected = 2/6  Using the law of total probability, we have   P(red ball) = P(A) = PE1×PAE1+PE2×PAE2                              = 12×47+12×26=1942

Answer: B) 19/90 Explanation: ASSISTANT→AAINSSSTT STATISTICS→ACIISSSTTT Here N and C are not common and same letters can be A, I, S, T. Therefore  Probability of choosing A =  2C19C1×1C110C1 = 1/45   Probability of choosing I = 19C1×2C110C1 = 1/45 Probability of choosing S = 3C19C1×3C110C1 = 1/10 Probability of choosing T = 2C19C1×3C110C1 = 1/15 Hence, Required probability =   145+145+110+115= 1990