BANK & INSURANCE (PERMUTATION AND COMBINATION) PART 2Total Questions: 3021. In a group of 7 boys and 9 girls, 5 members are to be selected. In how many different ways can they be selected such that at least one boy should be there?(a) 2450 (b) 4242 (c) 1840(d) 4280 (e) None of theseCorrect Answer: (b) 4242 Solution:Required number of ways => (7C4 and 9C1) or (7C3 and 9C2) or (7C2 and 9C3) or (7C1 and 9C4) or 7C5=> (35 × 9) + (35 × 36) + (21 × 84) + (7 × 126) + 21 => 315 + 1260 + 1764 + 882 + 21 = 424222. 17 buses are running between two places Nagercoil and Madurai. In how many ways can a family go from Nagercoil to Madurai and return by a different bus?(a) 220 ways (b) 210 ways(c) 252 ways (d) 272 ways(e) None of theseCorrect Answer: (d) 272 waysSolution:They can go in any bus out of the total 17 buses. They return by different buses, hence they cannot comeback in the same bus. Hence they can return in 16 ways. Total number of ways = 17 × 16 = 272 ways23. In a party hall, 10 persons are to be arranged around a round table. If two particular persons are not to be seated side by side, then what is the total number of arrangements?(a) 9 × 10! (b) 7 × 8!(c) 35 × 7! (d) 40 × 6!(e) None of theseCorrect Answer: (b) 7 × 8!Solution:No. of ways to arrange 10 persons around the table = (10 − 1)! = 9! No. of ways in which 2 particular persons sit side by side = 8! × 2! Therefore, required no. of arrangements = 9! − (8! × 2!) = 9 × 8! − 8! × 2 × 1 = (9 − 2) × 8! = 7 × 8!24. When 3 fair dice are rolled simultaneously, in how many outcomes will at least one of the dice show 3?(a) 91 (b) 87 (c) 68(d) 69 (e) None of theseCorrect Answer: (a) 91 Solution:When 3 dice rolled => Number of outcomes = 6³ = 216 Number of outcomes in which none of the 3 dice show 3 = 5³ = 125 Required no. of outcomes = 216 − 125 = 9125. In a group of 6 girls and 5 boys, 3 members are to be selected. In how many different ways can they be selected such that at least one girl should be there?(a) 195 (b) 210 (c) 155(d) 180 (e) None of theseCorrect Answer: (c) 155Solution:The possibilities are, => (1 girl and 2 boys) or (2 girls and 1 boy) or (3 girls)Required number of ways => (⁶C₁ and ⁵C₂) + (⁶C₂ and ⁵C₁) + (⁶C₃) => [6 × (5 × 4)/(1 × 2)] + [(6 × 5)/(1 × 2)] × 5] + [(6 × 5 × 4)/(1 × 2 × 3)] => 60 + 75 + 20 = 15526. A teacher wants to select a boy out of 8 boys and a girl out of 7 girls for the writing competition. In how many ways can be select?(a) 42 (b) 60 (c) 28(d) 56 (e) 36Correct Answer: (d) 56 Solution:Select a boy out of 8 boys and a girl out of 7 girls = 8 × 7 Total ways = 5627. In how many different ways a group of 5 men and 7 women can be formed out of 8 men and 10 women?(a) 4240(b) 6380(c) 6720(d) 5860 (e) None of theseCorrect Answer: (c) 6720Solution:Required number of ways => ⁸C₅ and ¹⁰C₇ => [(8 × 7 × 6 × 5 × 4)/(5 × 4 × 3 × 2 × 1)] × [(10 × 9 × 8 × 7 × 6 × 5 × 4)/(7 × 6 × 5 × 4 × 3 × 2 × 1)] => 56 × 120 = 672028. In a bag contains 2 orange and 3 apples. If 2 fruits are selected, in how many ways that can be selected such that at least one is apple?(a) 8(b) 9 (c) 10(d) 11 (e) None of theseCorrect Answer: (b) 9 Solution: Required ways = ⁵C₂ − ²C₂ = 10 − 1 = 929. The bank manager forms a secret 2 – digit code from the numbers 0-9. But he set code as the first digit will not be 0 and the second number will not be even number. Then what are the possible ways to set the code?(a) 54 (b) 55 (c) 64(d) 45 (e) 50Correct Answer: (d) 45 Solution:First number will not be zero implies there are 9 possible way for digit one = ⁹C₁ And for second digit we have 5 possibilities = (1, 3, 5, 7, 9) = ⁵C₁ So the possible number of ways = 9 × 5 = 4530. How many words can be formed by using all the letters of the word “NISARGA” so that the vowels are never together?(a) 2640 (b) 1230(c) 2460 (d) 4920(e) None of theseCorrect Answer: (c) 2460 Solution:Required number of ways = (7!/2!) − (5! 3!/2!) = [(7 × 6 × 5 × 4 × 3 × 2 × 1)/(2 × 1)] − [(5 × 4 × 3 × 2 × 1)/(2 × 1)] = 2460Submit Quiz« Previous123