In how many different ways can the letters of the word 'READING' be a

IMO Ans = 120.
READING = 7 LETTERS, if we remove AEI, there are only 4 letters remaining.
1 AEI _ _ _ _ (No . of ways of arranging 4 letters in 4 spaces = 4p4)
2 _ AEI _ _ _ (No . of ways of arranging 4 letters in 4 spaces = 4p4)
3. _ _ AEI _ _ (No . of ways of arranging 4 letters in 4 spaces = 4p4)
4. _ _ _ AEI _ (No . of ways of arranging 4 letters in 4 spaces = 4p4)
5. _ _ _ _AEI (No . of ways of arranging 4 letters in 4 spaces = 4p4)

Total no. ways of doing the required task = 5 X4p4 = 5 X 4! = 5! = 120.

Please let me know whether i am correct or not.

Regards
Narayana Raju

A

We consider AEI 1 unit because they will always be together

Hi Raju ,

You found the ways while considering no other Letters are in between AEI, however question is just asking to find out the ways such that letter A comes before E and E before I
So resulting words can be formed such as
ARENIDG ARGENID RAEINGD (few examples)

That's why the QA - B (840 ways calculated as 7!/3! ) considering all possible situations in which A comes before E and E comes before I.

Hope this makes question more clear.




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no. of ways in which 7 letters can be arranged = 7!
no. of ways in which those 3 letters can be arranged = 3!

since these 3 letters will always have only one arrange ment
therefore total ways = 7!/3! = 840

Option B

why did they divide by 3 factorial ??

No of ways to arrange all the letters in READING = 7!
This contains all the permutation in which A < E < I or A < I < E or I < E < A etc.

No of ways to arrange A , E , I = 3! .
Only in one of these A will come before E & I and E before I.

Therefore 7!/3!

Could someone please explain why we are dividing by 3!?

My understanding is as below:

we can select 3 places from the given 7 in 7C3 ways.Now,in the selection,there in only 1 way to arrange A,E,I in the desired sequence.Remaining 4 letters can be arranged in 4! ways.

So answer should be: 7C3* 4!= 7!/3!=840

EgmatQuantExpert chetan2u am I correct in my reasoning?

Yes, you are correct.

Another way to look at it is that the 7 words can be arranged in 7! Ways..
But 3 of them can be arranged in just 1 way rather than 3! Ways.
Therefore 7!/3! Ways

READING ; 7 letters which can be arranged in 7! ways
we need to find ways READING can be written where sequence of AEI is there
AEIRDNG ; eg so we can put AEI as X
XRDNG ; it can be written in 5! ways and placement of X can be in following order

RXDNG
RDXNG
RDNXG
RDNGX

5!*4 : 120*4 ; 840
IMO B

READING: 7 alphabets and no repeats.
Without restriction, these alphabets can be arranged in 7! ways = 5040 ways (max, so Choice E is out)

If AEI (vowels) are glued together as one unit in a fixed order, these alphabets , (AEI)RNDG, can be arranged in 5! ways (min, so Choice A is out)

Now, back to the question.
we should always work from slot with most restriction to slot with no restriction.
There is a restriction on alphabets AEI, their order is fixed (not glued together, just that the order cannot change).
For these 3 alphabets, lets first choose and then arrange.
We have 7 slots to choose from in 7C3 ways and 1 way to arrange as their order is fixed.
So, total ways to organize these alphabets is 7C3 * 1 = 7C3

Now, lets look at other 4 alphabets RNDG.
No repeats and no restrictions.
So, these 4 alphabets can simply be arranged in 4! ways.

Therefore, Total ways= (ways comprising fixed order for AEI) and (ways for remaining 4 RNDG) = 7C3 * 4! = 840.

Hence, Ans B

Please tell me why my approach is incorrect

There are 4! ways of arranging R, D, N and G

Let [ ] be a space where we put in one of these consonants and let a _ be a space where a vowel can fit in.

The letters can be arranged as follows:

_ _ _ [ ] _ _ _ [ ] _ _ _ [ ] _ _ _ [ ] _ _ _

First I arrange the consonants in 4! ways. Now, we can pic any 3 of the 15 spaces to put in our vowels (which can be arranged in only one particular order).

Using this thought process, we get

total number of words = 4! * 15C3

total letters=7 and we have to select AEI from it in 7C3 ways and rest of the 4 letter can be arranged in 4! ways.
7*5*24=840

Use the combination to choose 3 positions out of 7 7C3 now there is only 1 valid order for this position where A comes before E and E before I.

7C3*1 =35

Now rest of the 4 letters can we arrange in 4! ways = 24

total ways = 840