This section covers permutations and combinations. Arranging Objects The number of ways of arranging n unlike objects in a line is n! (pronounced ‘n factorial’). n! = n × (n – 1) × (n – 2) ×…× 3 × 2 × 1 Example How many different ways can the letters P, Q, R, S be arranged? The answer is 4! = 24. This is because there are four spaces to be filled: _, _, _, _ The first space can be filled by any one of the four letters. The second space can be filled by any of the remaining 3 letters. The third space can be filled by any of the 2 remaining letters and the final space must be filled by the one remaining letter. The total number of possible arrangements is therefore 4 × 3 × 2 × 1 = 4!
n! . Example In how many ways can the letters in the word: STATISTICS be arranged? There are 3 S’s, 2 I’s and 3 T’s in this word, therefore, the number of ways of arranging the letters are: 10!=50 400 Rings and Roundabouts
When clockwise and anti-clockwise arrangements are the same, the number of ways is ½ (n – 1)! Example Ten people go to a party. How many different ways can they be seated? Anti-clockwise and clockwise arrangements are the same. Therefore, the total number of ways is ½ (10-1)! = 181 440 Combinations The number of ways of selecting r objects from n unlike objects is:  Example There are 10 balls in a bag numbered from 1 to 10. Three balls are selected at random. How many different ways are there of selecting the three balls? 10C3 =10!=10 × 9 × 8= 120 Permutations A permutation is an ordered arrangement.
nPr = n! . Example In the Match of the Day’s goal of the month competition, you had to pick the top 3 goals out of 10. Since the order is important, it is the permutation formula which we use. 10P3 =10! = 720 There are therefore 720 different ways of picking the top three goals. Probability The above facts can be used to help solve problems in probability. Example In the National Lottery, 6 numbers are chosen from 49. You win if the 6 balls you pick match the six balls selected by the machine. What is the probability of winning the National Lottery? The number of ways of choosing 6 numbers from 49 is 49C6 = 13 983 816 . Therefore the probability of winning the lottery is 1/13983816 = 0.000 000 071 5 (3sf), which is about a 1 in 14 million chance. Suppose that, out of #n# things, #r_1# are of first type, #r_2# are of second type, #r_3# are of third type,..., where #r_1+r_2+r_3+...=n.# Then, no. of possible distinct permutations is given by #(n!)/{(r_1!)(r_2!)(r_3!)...}# In our Example, there are total #8# letters in the word INFINITY , out of which, #3# letters are of one type (i.e., the letter I ), #2# are of second type (i.e., the letter N ) and the remaining #3# are (i.e., the letters F,T and Y) are each of #1# type. Thus, #n=8, r_1=3, r_2=2, r_3=r_4=r_5=1#. #"The Reqd. No. of Permutations="(8!)/{(3!)(2!)(1!)(1!)(1!)}# #=(8xx7xx6xx5xx4)/(2!)=3360.# Enjoy Maths.!
Find the number of permutations of the letters of the word ‘PENDULAM', such that vowels are never together. (Approach 1)I know one solution is to find the permutation that vowels are together (ie; 6!) and subtract it from total number of permutations for the word (ie; 8!). Approach 2: I came across another interesting approach like shown below : Consonants in the word = $5$ ie;{P,N,D,L,M} Ways of arranging $5$ letters = $5!$ Consider a case like PNDLM and we insert positions that satisfy the condition of vowels not being together like _P_N_D_L_M_. So the vowels can be in any of the $6$ dashed positions (which is a equivalent to choosing $3$ out of $6$ positions ie; $6C3$) and vowels for chosen $3$ positions can be arranged among themselves in $3!$ ways. So the number of permutations should be : $6C3 \cdot 3! \cdot 5! = 14400$ (using counting principle) whereas solution by Approach 1 would be $= 8! - 6! = 39600$. Can someone point out where Approach 2 went wrong. I would like to use Approach 2 so that I can solve cases like
Improve Article Save Article Like Article Permutation is known as the process of organizing the group, body, or numbers in order, selecting the body or numbers from the set, is known as combinations in such a way that the order of the number does not matter. In mathematics, permutation is also known as the process of organizing a group in which all the members of a group are arranged into some sequence or order. The process of permuting is known as the repositioning of its components if the group is already arranged. Permutations take place, in almost every area of mathematics. They mostly appear when different commands on certain limited sets are considered. Permutation Formula In permutation r things are picked from a group of n things without any replacement. In this order of picking matter.
Combination A combination is a function of selecting the number from a set, such that (not like permutation) the order of choice doesn’t matter. In smaller cases, it is conceivable to count the number of combinations. The combination is known as the merging of n things taken k at a time without repetition. In combination, the order doesn’t matter you can select the items in any order. To those combinations in which re-occurrence is allowed, the terms k-selection or k-combination with replication are frequently used. Combination Formula In combination r things are picked from a set of n things and where the order of picking does not matter.
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Similar Questions Question 1: In how many ways can the letters be arranged so that all the vowels came together word is CORPORATION? Solution:
Question 2: In how many different ways can the letters of the word ‘MATHEMATICS’ be arranged such that the vowels must always come together? Solution:
Question 3: In How many ways the letters of the word RAINBOW be arranged in which vowels are never together? Solution:
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How many distinct permutations can the word probability have if the vowels should always come together?
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