What is the probability that the rearrangement of the word amazed the letter is positioned in between the two years?

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.

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)

Answer & Solution

Answer: Option 3

Solution:

The word AMAZED has six letters. Thus, there are six ways to place the M, five ways to place the Z, and four ways to place the D. Once those three letters have been placed, there is one way to fill the remaining positions with the letters A, E, and A, namely place the E in the middle open position and place the A's in the remaining positions. Therefore, there are $6 \cdot 5 \cdot 4 \cdot 1 = 120$ possible arrangements of the letters of the word AMAZED in which the E is positioned between the two A's.

Alternate Method: There are $\binom{6}{3}$ ways to choose the positions where the letters A, E, and A will go, one way to place them in that order in the chosen positions, and $3!$ ways to place M, Z, and D in the three remaining positions. Thus, there are $$\binom{6}{3} \cdot 3! = \frac{6!}{3!3!} \cdot 3! = \frac{6!}{3!} = 6 \cdot 5 \cdot 4 = 120$$ ways to arrange the letters of the word AMAZED so that the letter E is positioned between the two A's.

A nice alternative to the methods presented above can be found in the solutions provided by Federico Poloni and Angad.