What is the probability that in the rearrangements of the word amazed the letter E is positioned in between?

TCS Company Numerical Ability Permutation and Combination

• sorry its not necessary to be flanked (AEA)MZD=4!=24 (AEMA)ZD=3!*2!=12 (AEMDA)Z=2!*3!=12 (AEMDZA)=4!=24

  72 ANS

• 10 years agoHelpfull: Yes(24) No(8)
• ans:3)120 sol: at the 1st and last(sixth) place E can't be placed for 2nd and 5th place E(based on condition)....2*4!=48 for 3rd and 4th place E........................2*(3*3*2*2)=72 or 2*[(4!-3!)*2!]=72 (in other way can do this but both wil lgive same ans 72) so total arrangements=48+72=120
• 10 years agoHelpfull: Yes(13) No(9)
• In any rearrangement of the word, consider only the positions of the letters A, A and E. These can be as A A E, A E A or E A A. So, effectively one-third of all words will have 'E' in between the two 'A's. The total number of rearrangements are = 360. One-third of 360 is 120.
• 10 years agoHelpfull: Yes(9) No(3)
• bcz flank is not necessary so only AEA will not consider. there may be AMEAZD. there is also E is between 2A's.

  so ans will be 120

• 10 years agoHelpfull: Yes(7) No(7)
• (aea)zmd 4!=24
  
• 10 years agoHelpfull: Yes(5) No(8)
• we would place aea in a group and rest will be arranged as independent entities
  so the ans is 4!
• 10 years agoHelpfull: Yes(4) No(5)
• if anyone can elaborate more on this???
• 10 years agoHelpfull: Yes(3) No(0)
• According to the question E positioned between 2 A take AEA as 1 part. then total 4 parts. then there will be 4 places. place 1 can be arranged in 4 ways place 2 can be arranged in 3 ways place 3 can be arranged in 2 ways place 4 can be arranged in 1 ways

  so total 4! i.e. 4*3*2*1 = 24

• 10 years agoHelpfull: Yes(3) No(1)
• 120 (aea)zmd=48 azmeda=48 zameda=12 amedaz=12 total=120
• 10 years agoHelpfull: Yes(2) No(3)
• ans is 4*6=24
• 10 years agoHelpfull: Yes(2) No(4)
• in this question E should be in between two A like 'AEA' considered it as a single element then total no.of alphabates is 4 it can be arranged 4!=24 nd AEA can be arranged 3!=6 in which A come two time so 6/2!=3

  now total arrangement is 24*3=72 ans.

• 10 years agoHelpfull: Yes(2) No(1)
• please elaborate it na unable to catch it
• 10 years agoHelpfull: Yes(0) No(0)
• friend's it should be 4!
• 10 years agoHelpfull: Yes(0) No(2)
• (AEA)taken as 1 so total letters in (AEA)ZMD are 4
  combinations possible are 4!=24.
• 10 years agoHelpfull: Yes(0) No(2)
• take AEA as a unit so total way = 4!=24
• 10 years agoHelpfull: Yes(0) No(1)
• keep aea as one set and the terms aea,m,z,d can be arranged in 4! ways...so answer is 24...
• 10 years agoHelpfull: Yes(0) No(0)

The question is from Permutation and Combination. This is about Rearrangement of Letters. Our task is to find the probability of a word satisfying particular condition. This section hosts a number of questions which are on par with CAT questions in difficulty on CAT Permutation and Combination, and CAT Probability.

Question 18: If all the rearrangements of the word AMAZON are considered, what is the probability that M will feature between the 2As?

1. \\frac{1}{3}\\)
2. \\frac{1}{6}\\)
3. \\frac{2}{5}\\)
4. \\frac{3}{8}\\)

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For this type of question, we need to consider only the internal arrangement within the M and 2As. M and 2As can be rearranged as AMA, AAM, or MAA. So, the probability that M will feature between the 2As is \\frac{1}{3}\\). Now, let us think why we need to consider only the M and 2As. Let us start by considering a set of words where the M and 2 As are placed at positions 2, 3 and 5. The other three letters have to be in slots 1, 4 and 6 Three letters can be placed in three different slots in 3! = 6 ways. Now with __ M A __ A __ there are 6 different words. With __ A M __ A __ there are 6 different words. With __ A A __ M __ there are 6 different words. For each selection of the positions for A,A and M, exactly one-third of words will have M between the two A’s. This is why only the internal arrangement between A, A and M matters. So, probability of M being between 2 As is \\frac{1}{3}\\).

The question is "what is the probability that M will feature between the 2As?"

Hence the answer is "\\frac{1}{3}\\)"

Choice A is the correct answer.

Answer & Solution

Answer: Option 3

Solution: