How many ways can the letters of the word Arrange be arranged so that two Rs come together?

$\begingroup$

In how many ways can the letters of the word ARRANGE be arranged such that two R's are never together?

Now total number of words are $\frac{7!}{2!2!}$.

Now words in which R's are always together are $\frac{6!}{2!2!}$. Subtracting these two doesn't give me correct answer. Can you explain why?

Thanks.

asked Apr 11, 2017 at 5:41

$\endgroup$

1

$\begingroup$

In your second case. It will be just $\frac{6!}{2!}$. Because your are taking 'RR' as a single unit which can be adjusted only one way

answered Apr 11, 2017 at 5:43

How many ways can the letters of the word Arrange be arranged so that two Rs come together?

The Dead LegendThe Dead Legend

4,6871 gold badge15 silver badges36 bronze badges

$\endgroup$

8

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

Verified

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.

How many ways can the letters of the word Arrange be arranged so that 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.

How many ways can the letters in the word parrot can be arranged if R's are to be seperated then find the number of arrangements?

Required number of ways = 60 × 2 = 120 ways. ∴ "PARROT" can be arranged in such a way that the vowels always come together for that we have 120 ways.

How many different ways can the letters of the word be arranged?

When the vowels OIA are always together, they can be supposed to form one letter. Then, we have to arrange the letters PTCL (OIA). Now, 5 letters can be arranged in 5! = 120 ways. ... Exercise :: Permutation and Combination - General Questions..

How many different ways are there to arrange letters of the world how many of these arrangements begin with the letter R?

of arrangements of the word=120. No. of arrangements taking R at beginning=24.