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My name is AJEET. But my son accidentally types the name by
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10 Nov 2007, 21:53
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53% (01:38) correct 47% (01:42) wrong based on 472 sessions
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My name is AJEET. But my son accidentally types the name by interchanging a pair of letters in my name. What is the probability that despite this interchange, the name remains unchanged? A. 5% B. 10% C. 20% D. 25%
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Re: probability
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26 Aug 2008, 13:51
Balvinder wrote: 115. My name is AJEET. But my son accidentally types the name by interchanging a pair of letters in my name. What is the probability that despite this interchange, the name remains unchanged?
a) 5% b) 10% c) 20% d) 25% = (No of ways to chose two letter for exchangefrom{EE}) /(No of ways to chose two letter for exchange from{AJEET}) =2C2/5C2 = 1/10=10%
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Joined: 25 Jul 2007
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This is a very crude method but i guess it's effective for such small problems.
Possible pairs of letters that could be interchanged are:
AJ
AE
AE
AT
JE
JE
JT
EE
ET
ET
Out of the the ten possibilities, only 1 would lead to an unchanged name.
Therefore, the probability is 1/10 = 10 %.
A more general approach goes as follows:
Each letter can be interchanged in the following ways.
'A' can be interchanged with 4 letters,' 'J' with 3, 'E' with 2 and 'E' with 1.
Therefore total possibilities equal 4+3+2+1=10.
Again only 1 would lead to an unchanged name.
Therefore the answer is 1/10 =10%



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Joined: 24 Sep 2005
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jbs wrote: This is a very crude method but i guess it's effective for such small problems.
Possible pairs of letters that could be interchanged are:
AJ AE AE AT JE JE JT EE ET ET
Out of the the ten possibilities, only 1 would lead to an unchanged name. Therefore, the probability is 1/10 = 10 %.
A more general approach goes as follows:
Each letter can be interchanged in the following ways.
'A' can be interchanged with 4 letters,' 'J' with 3, 'E' with 2 and 'E' with 1.
Therefore total possibilities equal 4+3+2+1=10.
Again only 1 would lead to an unchanged name.
Therefore the answer is 1/10 =10%
It's nice
BTW, I think picking a pair to interchange its members is like picking 2 out of 5 letters, thus C(5,2) ( We dont care their order anyways)



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Re: probability
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26 Aug 2008, 17:32
whts the OA ?
i think answer shud be 20%
i divided the 5C2 by 2 as there are 2 E's and hence pairs wud get repeated but the Name wudnt change
as in even if the boy replaces A with first E or the second E , the Name remains the same .



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Re: probability
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26 Aug 2008, 17:34
ok i think i got where i made the mistake answer shud be 10%



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Re: probability
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28 Sep 2009, 10:45
115. My name is AJEET. But my son accidentally types the name by interchanging a pair of letters in my name. What is the probability that despite this interchange, the name remains unchanged?
a) 5% b) 10% c) 20% d) 25%
Soln: The total number of ways in which two letters can be interchanged and we get a new five letter words is 10 ways.
Of these 10 possible five letter words, the only interchange where the name does not change is when the E's are interchanged.
Thus probability is = (1/10) * 100 = 10%
Ans is B



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Re: probability
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08 Jan 2010, 21:56
is my approach correct? I read the question as, what is the prob. to choose EE from AJEET, then, 2/5*1/4=1/10 ==> 10% ? thanks.



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Re: probability
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01 Nov 2013, 06:33
bekbek wrote: is my approach correct? I read the question as, what is the prob. to choose EE from AJEET, then, 2/5*1/4=1/10 ==> 10% ? thanks. Hey thats a very good way of solving and in fact the easiest and fastest way with the least probability of err



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Re: probability
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12 Nov 2013, 09:37
bekbek wrote: is my approach correct? I read the question as, what is the prob. to choose EE from AJEET, then, 2/5*1/4=1/10 ==> 10% ? thanks. HI There I didnt quite understand your calculation although I took the a similar logic I read the question as " what is the probability of EE being together in AJEET " so combinations where EE is together in AJEET is 4!/2! (Club EE together as 1 unit, hence AJ(EE)T, different combos => 4! and EE repeats twice so divide by 2!) probability of EE together in all combinations of AJEET = (4!/2!)/5! * 100 => 12/120 * 100 => 10% Is this method correct??



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Re: My name is AJEET. But my son accidentally types the name by
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18 Feb 2014, 06:25
We need to pick 2 letters from the 5 given. As only one combination of letters will make the name remain the same, that is, changing the order of E and E. then 1/5C3 = 1/10 = 10%. B is the correct answer
Hope this helps Cheers J



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Re: My name is AJEET. But my son accidentally types the name by
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07 May 2016, 04:03
Hi Bunuel, please give a solution to this problem. Thanks.



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Re: My name is AJEET. But my son accidentally types the name by
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07 May 2016, 04:14
Proba = (Number of pair (E,E))/(total number of pairs)
Number of pair (E,E) = 1
Total number of pairs = (5* 4)/2 = 10
So, Proba = 1/10 = 10%



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Re: My name is AJEET. But my son accidentally types the name by
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07 May 2016, 04:20
shalinkotia wrote: Hi Bunuel, please give a solution to this problem. Thanks. Hi, the name is AJEET.. Any two letters are changed.. total letters = 5 and we choose two out of these = \(5C2 = \frac{5!}{3!2!} = 10\).. there is ONLY 1 combination where the name does not change E with E.. 1 way \(prob = \frac{1}{10} = 10%\)
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Re: My name is AJEET. But my son accidentally types the name by
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18 May 2016, 13:02
Shouldn't the total number of ways of selecting 2 letters out of 5 be 4c2(4 disitnct letters A J E T)+1(EE) Bunuel please help?



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Re: My name is AJEET. But my son accidentally types the name by
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13 Aug 2016, 17:57
4c2(4 disitnct letters A J E T)+1(EE) 
Order matters, so I don't think you shouldn't be using a combination. Here's a refresher: Combinations: When the outcome of each stage does not differ from the outcomes of the others Permutations: When order matters
Permutation formula: \(\frac{n!}{(nr)!}\)
Applying the formula to the question yields: \(\frac{5!}{(52)!} = \frac{5!}{3!}\) = 20 total outcomes
There are two E's in the name, so there are 2 outcomes out of 20 that the son could select. Thus, \(\frac{2}{20} = \frac{1}{10}\) = 10%.



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Re: My name is AJEET. But my son accidentally types the name by
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14 Jun 2017, 00:51
here i calculated the total ways of arranging the word AJEET which is 60 because it is 5!/ 2! = 60 why is this wrong??



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Re: My name is AJEET. But my son accidentally types the name by
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17 Aug 2019, 10:11
Balvinder wrote: My name is AJEET. But my son accidentally types the name by interchanging a pair of letters in my name. What is the probability that despite this interchange, the name remains unchanged?
A. 5% B. 10% C. 20% D. 25% There are 5C2 ways of interchanging 2 letters which is 10. Out of those 10 ways, the only way the name will remain intact is when the son will pick EE as 2 letters. 1/10. Hence 10%. It clicked me so I took 26 seconds to solve this one. Lol Thank you!
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Re: My name is AJEET. But my son accidentally types the name by
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18 Aug 2019, 05:08
Hello. I still didn't understand why we have 10 possible outcomes instead of 5 i.e; 5C2/2!, Since there are 2 Es. Could anyone pls help me understand? TIA
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Re: My name is AJEET. But my son accidentally types the name by
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