Permutations And Combinations Sample Questions And Answers
Ex. 1. Evaluate: 30!/28! Sol. We have, 30!/28! = 30x29x(28!)/28! = (30x29) = 870.Ex. 2. Find the value of (i)60p3 (ii) 4p4 Sol. (i) 60p3 = 60!/(60-3)! = 60!/57! = 60x59x58x(57!)/57! = (60x59x58) = 205320. (ii) 4p4 = 4! = (4x3x2x1) = 24.Ex. 3. Find the vale of (i)10c3 (ii) 100c98 (iii)50c50 Sol. (i) 10c3 = 10x9x8/3! = 120. (ii) 100c98 = 100c(100-98)= 100x99/2! = 4950. (iii) 50c50 = 1. [ncn = 1]Ex. 4. How many words can be formed by using all letters of the word 'BIHAR' Sol. The word BIHAR contains 5 different letters.Required number of words = 5p5 = 5! = (5x4x3x2x1) = 120.Ex. 5. How many words can be formed by using all letters of the word 'DAUGHTER' so that the vowels always come together? Sol. Given word contains 8 different letters. When the vowels AUE are always together, we may suppose them to form an entity, treated as one letter.Then, the letters to be arranged are DGNTR (AUE).Then 6 letters to be arranged in6p6 = 6! = 720 ways.The vowels in the group (AUE) may be arranged in 3! = 6 ways.Required number of words = (720x6) = 4320.Ex. 6. How many words can be formed from the letters of the word 'EXTRA' so that the vowels are never together? Sol. The given word contains 5 different letters. Taking the vowels EA together, we treat them as one letter. Then, the letters to be arranged are XTR (EA). These letters can be arranged in 4! = 24 ways. The vowels EA may be arranged amongst themselves in 2! = 2 ways. Number of words, each having vowels together = (24x2) = 48 ways. Total number of words formed by using all the letters of the given words = 5! = (5x4x3x2x1) = 120.Number of words, each having vowels never together = (120-48) = 72.Ex. 7. How many words can be formed from the letters of the word 'DIRECTOR'So that the vowels are always together? Sol. In the given word, we treat the vowels IEO as one letter. Thus, we have DRCTR (IEO). This group has 6 letters of which R occurs 2 times and others are different. Number of ways of arranging these letters = 6!/2! = 360. Now 3 vowels can be arranged among themselves in 3! = 6 ways. Required number of ways = (360x6) = 2160.Ex. 8. In how many ways can a cricket eleven be chosen out of a batch of15 players ? Sol. Required number of ways = 15c11 =15c(15-11) = 11c4 = 15x14x13x12/4x3x2x1 = 1365.Ex. 9. In how many ways, a committee of 5 members can be selected from6 men and 5 ladies, consisting of 3 men and 2 ladies? Sol. (3 men out 6) and (2 ladies out of 5) are to be chosen.Required number of ways = (6c3x5c2) = [6x5x4/3x2x1] x [5x4/2x1] = 200.
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