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

Question

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

firas92 · Answer

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}

Answer is  (C)

NitishJain · Answer

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

Eduardo86 · Answer

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?

ScottTargetTestPrep · Answer

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:

.

Thus, the probability of this occurrence is:

/

= 1 -  /  -  /

= 1 - 10!//11! -  /

= 1 - 1//11 -  /11!

= 1 - 1/11 -  /11

= 1 - 3/11 = 8/11

Answer: C

Taulark1 · Answer

Can someone please explain the part where - Number of permutations with E as first alphabet = 10!2! ??

Kritisood · Answer

=> 1 - prob( 1st letter is T or E)
1 - (1/11+2/11)
1-3/11
8/11

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

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

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