BANK & INSURANCE (PERMUTATION AND COMBINATION) PART 1Total Questions: 3011. In how many different ways can the letters of the word “CABINET” be arranged?(a) 5040 (b) 2520 (c) 720(d) 240 (e) 360Correct Answer: (a) 5040 Solution:Required ways = 7! = 504012. A teacher and head master are chosen out of a group having 15 persons. How many ways are there?(a) 120 (b) 180 (d) 240 (c) 210(e) 280Correct Answer: (d) 240 Solution:Total number of ways = 15 × 14 = 21013. In how many ways the letters of the word ‘ARMOUR’ can be arranged?(a) 720 (b) 360 (c) 120(d) 650 (e) None of TheseCorrect Answer: (b) 360 Solution:Total letters = 6, but R has come twice. No of arrangements = 6!/2! = (6 × 5 × 4 × 3 × 2 × 1)/(2 × 1) = 36014. How many four letter words can be formed out of the letters of the word is “LOGARITHMS”?(a) 2520 (b) 720 (c) 5040(d) 360 (e) None of theseCorrect Answer: (c) 5040Solution:Number of letters = 10!/(10 − 4)! = 720 × 7 = 504015. In how many ways word “BANKING” can be arranged in such a way that all vowels and consonants comes together?(a) 120 (b) 280 (c) 360(d) 240 (e) None of theseCorrect Answer: (d) 240 Solution:Vowels = A, I = 2! Consonant = B, N, K, N, G = 5! Required ways = 2! × 5!/2! × 2! = 24016. In a class has 3 boys and 2 girls, two students were selected. In how many different ways can they be selected such that at least one girl should be there?(a) 5 (b) 8(c) 7(d) 6 (e) None of theseCorrect Answer: (c) 7Solution:Total number of ways = ⁵C₂ No girls selected = ³C₂ Required ways = ⁵C₂ − ³C₂ = 10 − 3 = 717. Two students are selected from 8 students. How many ways are there to achieve this?(a) 68 (b) 64(c) 56(d) 52 (e) None of theseCorrect Answer: (c) 56Solution:Number of ways = 8 × 7 = 5618. In how many ways the letters of the word “COURSE” can be arranged?(a) 360 (b) 240 (c) 540(d) 128 (e) None of theseCorrect Answer: (e) None of theseSolution:Required ways = 6! = 72019. A bag contains 4 red balls and three green balls. If two balls are selected, in how many different ways the balls selected such that at least one red ball should be there?(a) 18 (b) 20 (c) 36(d) 40 (e) None of theseCorrect Answer: (a) 18 Solution:Total number of ways = ⁷C₂ No red balls = ³C₂ Required ways = ⁷C₂ − ³C₂ = 7 × 6/2 − 3 × 2/2 = 36/2 = 1820. How many three digit numbers can be formed with the digits 2, 3, 5, 6, 9, if repetition of digits is allowed?(a) 120 (b) 240 (c) 144(d) 720(e) None of theseCorrect Answer: (e) None of theseSolution:The number of ways in three digit no’s can be formed from the given digit is 5 × 5 × 5 = 125Submit Quiz« Previous123Next »