In how many ways can be the letters of the word ARRANGE be arranged as per the following cases: (a) The two R’s are never together (b) The two A’s are together but not two R’s (c) Neither two A’s nor the two R’s are together.
ANSWER
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Hint: For solving this question few will use the formula of the different number of arrangements of $n$ objects out of which few objects are of similar types. Then, we will try to find the answer to each part.
Complete step by step answer: Given: The word ARRANGE. It has a total of 7 words out of which there are two A’s and two R’s and the rest three words are different. Now, when we have $n$ objects such that $p$ items are of one type and $q$ are of another type and the rest $r$ are different objects, so $n=p+q+r$ . Then, we can arrange them in $\dfrac{n!}{\left( p! \right)\left( q! \right)}$ ways. Now, in our question, we have word ARRANGE where total 7 words are there out of which 2 are A’s and 2 are R’s and rest 3 are different then, the total number of ways in which letters of the given word can be arranged will be $\dfrac{7!}{\left( 2! \right)\left( 2! \right)}=1260$ . (a) The two R’s are never together: Now, for this case we will club the two R’s together and treat it as one single letter then we will have a total 6 words out of which 2 are A’s and rest 4 are of different types. And we can arrange them in $\dfrac{6!}{\left( 2! \right)}=360$ the number of ways. Then, when we subtract this from the total number of arrangements that is 1260 then we will get the number of arrangements in which the two R’s are never together. Thus, we can arrange the letter of the word ARRANGE such that two R’s are never together is equal to $1260-360=900$ . Hence, 900 such arrangements are possible in which two R’s are never together. (b) The two A’s are together but not two R’s: Now, for this case we will club the two A’s together and treat it as one single letter then we will have a total 6 words out of which 2 are R’s and the rest 4 are of different types. And we can arrange them in $\dfrac{6!}{\left( 2! \right)}=360$ the number of ways. Then, the number of ways in which 2 A’s are together is 360. Now, club the two A’s and treat it as one single letter and the two R’s and treat it as one single letter then we will have a total 5 different words which can be arranged in $5!=120$ ways. Then, the number of ways in which two A’s and two R’s both are together is 120. Now, when we subtract the number of ways in which two A’s and two R’s both are together from the number of ways in which 2 A’s are together then we will get the number of ways in which the two A’s are together but not two R’s and that is $360-120=240$ ways. Hence, 240 such arrangements are possible in which two A’s are together but not two R’s. (c) Neither two A’s nor the two R’s are together. Now, when we subtract the number of ways in which the two A’s are together but not two R’s from the number of arrangements in which the two R’s are never together then we will get the number of arrangements in which neither two A’s nor the two R’s are together and that is $900-240=660$ ways. Hence, 660 such arrangements are possible in which neither two A’s nor the two R’s are together.
Note: Here, the student must take care while making different cases that are possible and not directly apply the formula for the total number of different arrangements of a certain number of different objects in a linear arrangement.