![]() ![]() Q.7: Six points are marked on a straight line and five points are marked on another line which is parallel to the first line. SSC Exam Test and Preparation 2023 – Download Now! More than 1200+ detailed question and answers Prepare for SSC CGL, CHSL, SSC MTS, Junior Engineer 2023 Q.6: The number of new words that can be formed by rearranging the letters of the word 'ALIVE' is:Įxplanation: Number of words which can be formed = 5! – 1 = 120 – 1 = 119 How many second class tickets have to be printed, so that a passenger can travel from any station to any other station?Įxplanation: The total number of stations = 20įrom 20 stations we have to choose any two stations and the direction of travel (i.e., Hyderabad to Bangalore is different from Bangalore to Hyderabad) in 20P 2 ways. Q.5: There are 18 stations between Hyderabad and Bangalore. Permutation – Basic Concepts, Methods, Examples, Math Tricks to Solve (Part 2) (Quantitative Aptitude made Simpler) Using these letters the number of 8 letters words formed is 8P 8 = 8! Q.4: Using all the letters of the word 'THURSDAY', how many different words can be formed? The number of 4 digit numbers that can be formed using six digits is 6P 4 = 6 * 5 * 4 * 3 = 360 Q.3: How many 4 digit numbers can be formed using the digits (1, 3, 4, 5, 7, 9) when repetition of digits is not allowed? Permutation – Basic Concepts, Methods, Examples, Math Tricks to Solve (Part 1) (Quantitative Aptitude made Simpler) Number of ways such that two ladies are always included in the committee = 6C 3 = (6 * 5 * 4) / 6 = 20 In how many ways the delegation can be formed, if 2 particular ladies are always included in the delegation?Įxplanation: There are three ladies and five gentlemen and a committee of 5 members to be formed. Q.2: A delegation of 5 members has to be formed from 3 ladies and 5 gentlemen. To find how many 4 letter word we can find from that = 10*9*8*7 = 5040 Questions and Solved Examples on Permutation and Combination Q.1: How many 4-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?Įxplanation: The Word LOGARITHMS contain 10 letters. Number of diagonals in a geometric figure of n sides = nC 2-n.There are 60 different arrangements of these letters that can be made. Finally, when choosing the third letter we are left with 3 possibilities. After that letter is chosen, we now have 4 possibilities for the second letter. For the first letter, we have 5 possible choices out of A, B, C, D, and E. Let us break down the question into parts. \( \Longrightarrow \) There are 60 different arrangements of these letters that can be made. \( \Longrightarrow\ _nP_r =\ _5P_3 = 60 \) applying our formula \( \Longrightarrow r = 3 \) we are choosing 3 letters \( \Longrightarrow n = 5 \) there are 5 letters Let us first determine our \( n \) and \( r \): We will solve this question in two separate ways. If the possible letters are A, B, C, D and E, how many different arrangements of these letters can be made if no letter is used more than once? ![]() When dealing with more complex problems, we use the following formula to calculate permutations:Ī football match ticket number begins with three letters. The arrangements of ACB and ABC would be considered as two different permutations. Suppose you need to arrange the letters A, C, and B. \( \Longrightarrow \) There are 10 ways in which Katya can choose 3 different cookies from the jar.Īs mentioned in the introduction to this guide, permutations are the different arrangements you can make from a set when order matters. \( \Longrightarrow\ _nC_r =\ _5C_3 = 10 \) applying our formula \( \Longrightarrow r = 3 \) we are choosing 3 cookies \( \Longrightarrow n = 5 \) there are 5 cookies Since order was not included as a restriction, we see that this is a combination question. We must first determine what type of question we are dealing with. In how many ways can Katya choose 3 different cookies from the jar? Katya has a jar with 5 different kinds of cookies. ![]() Where \( n \) represents the total number of items, and \( r \) represents the number of items being chosen at a time. When dealing with more complex problems, we use the following formula to calculate combinations: The arrangements of ACB and ABC would be considered as one combination. As introduced above, combinations are the different arrangements you can make from a set when order does not matter. ![]()
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