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My name is AJEET. But my son accidentally types the name by [#permalink]

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10 Nov 2007, 21:53

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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?

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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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.

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%

Re: My name is AJEET. But my son accidentally types the name by [#permalink]

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18 Feb 2014, 06:25

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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

Re: My name is AJEET. But my son accidentally types the name by [#permalink]

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02 May 2015, 06:02

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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 [#permalink]

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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: \(\frac{n!}{(n-r)!}\)

Applying the formula to the question yields: \(\frac{5!}{(5-2)!} = \frac{5!}{3!}\) = 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%.