How many words can be formed using the letters of the word arrow where two rs do not come together

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The word ARRANGEMENT has $11$ letters, not all of them distinct. Imagine that they are written on little Scrabble squares. And suppose we have $11$ consecutive slots into which to put these squares.

There are $\dbinom{11}{2}$ ways to choose the slots where the two A's will go. For each of these ways, there are $\dbinom{9}{2}$ ways to decide where the two R's will go. For every decision about the A's and R's, there are $\dbinom{7}{2}$ ways to decide where the N's will go. Similarly, there are now $\dbinom{5}{2}$ ways to decide where the E's will go. That leaves $3$ gaps, and $3$ singleton letters, which can be arranged in $3!$ ways, for a total of $$\binom{11}{2}\binom{9}{2}\binom{7}{2}\binom{5}{2}3!.$$

1) 360

2) 900

3) 1260

4) 1620

Answer: (2) 900

Solution: In the given word “ARRANGE”, we have two R’s, two A’s and rest letters are one each.

Total number of ways = 7!/2!.2! = 1260

Number of ways when R’s are together = 6!/2! = 360

Therefore, number of ways when both R do not come together = 1260 – 360 = 900

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Text Solution

Solution : The letters of word ARRANGE can be rewritten as A R N G E A R So we have 2 A's and 2 R's , and total 7 letters. (i) Total number of words is `(7!)/(2!2!)=1260`. The number of words in which 2 R's are together [consider (R R) as one unit] is `6!//2!`. e.g., The number of words in which 2 R's are together [consider (R R) as one unit] is `6!//2!`. e.g., (R R),A,A,N,G,E Note that permutations of R R give nothing extra. Therefore, the number of words in which the two R's are never together is `(7!)/(2!2!)-(6!)/(2!)=900` (ii) The number of words in which both A's are together is `6!//2!=360`, e.g., (A A),R,R,N,G,E The number of words in which both A's and both R's are together is 5!=120, e.g., (A A), (R R), N,G,E Therefore, the number of words in which both A's are together but the two R's are not together is 360-120=240. (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.