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# What is the probability that when the letters of the word SERENDIPITY

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Math Expert
Joined: 02 Sep 2009
Posts: 64949
What is the probability that when the letters of the word SERENDIPITY  [#permalink]

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02 Jun 2020, 06:52
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Difficulty:

25% (medium)

Question Stats:

80% (01:55) correct 20% (01:52) wrong based on 64 sessions

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What is the probability that when the letters of the word SERENDIPITY are randomly rearranged, the first alphabet of the resulting word is neither T nor E?

A. 10/11
B. 9/11
C. 8/11
D. 5/11
E. 4/11

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Re: What is the probability that when the letters of the word SERENDIPITY  [#permalink]

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02 Jun 2020, 11:08
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Bunuel wrote:
What is the probability that when the letters of the word SERENDIPITY are randomly rearranged, the first alphabet of the resulting word is neither T nor E?

A. 10/11
B. 9/11
C. 8/11
D. 5/11
E. 4/11

Total Number of permutations of the word SERENDIPITY $$= \frac{11!}{2!2!}$$

Number of permutations with T as first alphabet $$= \frac{10!}{2!2!}$$

Number of permutations with E as first alphabet $$= \frac{10!}{2!}$$

So the number of possible permutations with neither T nor E being the first alphabet $$= \frac{11!}{2!2!}-\frac{10!}{2!2!}-\frac{10!}{2!}$$ $$= \frac{10!}{2!}(\frac{11}{2}-\frac{1}{2}-1)$$ $$= \frac{10!}{2!}(4)$$

Therefore, the probability that when the letters of the word SERENDIPITY are randomly rearranged, the first alphabet of the resulting word is neither T nor E $$= \frac{\frac{10!}{2!}(4)}{\frac{11!}{2!2!}}$$

$$= \frac{\frac{10!}{2!}(4)}{\frac{10!}{2!}(\frac{11}{2})} = \frac{8}{11}$$

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Joined: 11 Feb 2019
Posts: 270
Re: What is the probability that when the letters of the word SERENDIPITY  [#permalink]

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03 Jun 2020, 13:14
1
IMO C

Total # of words = 11!/ (2! * 2!)

P(first alphabet of the resulting word is neither T nor E) = 1 - P(first alphabet of the resulting word is T or E)

P(first alphabet of the resulting word is T) = (10!/2!*2!) / (11!/2!*2!) = 1/11
P(first alphabet of the resulting word is E) = (10!/2!) / (11!/2!*2!) = 2/11

P(first alphabet of the resulting word is neither T nor E) = 1 - (1/11 + 2/11) = 8/11
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Re: What is the probability that when the letters of the word SERENDIPITY  [#permalink]

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04 Jun 2020, 22:32
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My strategy was first to calculate the probability to have T and E

p(T) is 1/11 and P(E) 2/11

Then my second step was to calculate the P to have both T and E as the first letter, 1/11 + 2/11 = 3/11

To find the final value I have subtract 1 - 3/11 = 8/11

Can anyone confirm if my strategy was correct or not?
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Re: What is the probability that when the letters of the word SERENDIPITY  [#permalink]

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06 Jun 2020, 05:35
1
Bunuel wrote:
What is the probability that when the letters of the word SERENDIPITY are randomly rearranged, the first alphabet of the resulting word is neither T nor E?

A. 10/11
B. 9/11
C. 8/11
D. 5/11
E. 4/11

Solution:

We use the indistinguishable permutations formula to calculate the total number of ways to arrange the letters in the word SERENDIPITY: 11! / (2! x 2!). (Note that the two E’s and two I’s are indistinguishable, so we must divide 11! by 2! twice to account for this.)

If T is the first letter, then there are 10! / (2! x 2!) ways to arrange the remaining letters.

If (one of the) E is the first letter, then there are 10! / 2! ways to arrange the remaining letters.

We see that the total number of ways to arrange the letters, taking into account the stated restrictions, is:

[11! / (2! x 2!) - 10! / (2! x 2!) - 10! / 2!].

Thus, the probability of this occurrence is:

[11! / (2! x 2!) - 10! / (2! x 2!) - 10! / 2!] / [11!/(2! x 2!)]

= 1 - [10!/(2! x 2!)]/[11!/(2! x 2!)] - [10!/2!]/[11!/(2! x 2!)]

= 1 - 10!//11! - [10! x 2!/(2! x 2!)]/[11!/(2! x 2!)]

= 1 - 1//11 - [10! x 2!/]/11!

= 1 - 1/11 - [1 x 2]/11

= 1 - 3/11 = 8/11

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Re: What is the probability that when the letters of the word SERENDIPITY  [#permalink]

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08 Jun 2020, 07:16
Can someone please explain the part where - Number of permutations with E as first alphabet = 10!2! ??
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Re: What is the probability that when the letters of the word SERENDIPITY  [#permalink]

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02 Jul 2020, 12:39
=> 1 - prob( 1st letter is T or E)
1 - (1/11+2/11)
1-3/11
8/11
Re: What is the probability that when the letters of the word SERENDIPITY   [#permalink] 02 Jul 2020, 12:39