10 Given, the word ARRANGE. It has 7 letters of which 2 letters (A, R) are repeating. The letter A is repeated twice, and the letter R is also repeated twice in the given word. To find: Number of ways the letters of word ARRANGE be arranged in such a way that not all R’s do come together. First, we find all arrangements of word ARRANGE, and then we minus all those arrangements of word ARRANGE in such a way that all R’s do come together, from it. This will exactly be the same as- all number of arrangements such that not all R’s do come together. Since we know, Permutation of n objects taking r at a time is nPr,and permutation of n objects taking all at a time is n! And, we also know Permutation of n objects taking all at a time having p objects of the same type, q objects of another type, r objects of another type is . i.e. the, number of repeated objects of same type are in denominator multiplication with factorial.A total number of arrangements of word ARRANGE: Total letters 7. Repeating letters A and R, the letter A repeating twice and letter R repeating twice. The total number of arrangements Now we find a total number of arrangements such that all R’s do come together. A specific method is usually used for solving such type of problems. According to that, we assume the group of letters that remain together (here R, R) is assumed to be a single letter and all other letters are as usual counted as a single letter. Now find a number of ways as usual; the number of ways of arranging r letters from a group of n letters is equals to nPr. And the final answer is then multiplied by a number of ways we can arrange the letters in that group which has to be stuck together in it (Here R, R). Letters in word ARRANGE: 7 letters Letters in a new word: A, RR, A, N, G, E: 6 letters (Letter A repeated twice). Total number of word arranging all the letters where second fraction 2! divided by 2! comes from arranging letters inside the group RR: arrangements of two letters where all the two letters are same = 2!/2! = 1 (Obviously! You can even think of it). Now, a Total number of arrangements where not all R’s do come together is equals to total arrangements of word ARRANGE minus the total number of arrangements in such a way that all R’s do come together. = 900 Hence, the total number of arrangements of word ARRANGE in such a way that not all R’s come together is equals to 900. Open in App 1 Text Solution Solution : The letters of word ARRANGE can be rewritten as <br> A R N G E <br> A R <br> So we have 2 A's and 2 R's , and total 7 letters. <br> (i) Total number of words is `(7!)/(2!2!)=1260`. <br> The number of words in which 2 R's are together [consider (R R) as one unit] is `6!//2!`. e.g., <br> The number of words in which 2 R's are together [consider (R R) as one unit] is `6!//2!`. e.g., <br> (R R),A,A,N,G,E <br> Note that permutations of R R give nothing extra. Therefore, the number of words in which the two R's are never together is <br> `(7!)/(2!2!)-(6!)/(2!)=900` <br> (ii) The number of words in which both A's are together is `6!//2!=360`, e.g., <br> (A A),R,R,N,G,E <br> The number of words in which both A's and both R's are together is 5!=120, e.g., <br> (A A), (R R), N,G,E <br> Therefore, the number of words in which both A's are together but the two R's are not together is 360-120=240. <br> (iii) There are in all 900 words in each of which the two R's are never together. Consider any such word. Either the two A's are together or the two A's not together. But the number of all such arrangements in which the two A's are together is 240. Hence, the number of all such arrangements in which the two A's not together is 900-240=660. |