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Re: From a class of 20 students whose names are listed in alphabetical. [#permalink]
where do you get 1140 from?
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Re: From a class of 20 students whose names are listed in alphabetical. [#permalink]
safety16644 wrote:
where do you get 1140 from?



The answer to your question is in the prior response

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Re: From a class of 20 students whose names are listed in alphabetical. [#permalink]
Yes, however I dont understand how they computed that number and arrived to 1140, hoping for a little bit more detail on it. 

20 C3 =1140 doesnt make sense to me, how did they arrive to 1140?
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From a class of 20 students whose names are listed in alphabetical. [#permalink]
safety16644 wrote:
Yes, however I dont understand how they computed that number and arrived to 1140, hoping for a little bit more detail on it. 

20 C3 =1140 doesnt make sense to me, how did they arrive to 1140?


OK.

You are asking a very basic question, no harm in that, but I would recommend studying the basics of combinatorics on this site and then trying some of the less difficult questions before attempting questions at this level.

1140 represents the number of different ways 3 students can be selected from 20 without restriction.

That result is obtained by the basic formula for combinations:

20!/3!17! =

(20*19*18*17...1)/(17*16*15...1)*(3*2*1)

Every multiplication from 17 to 1 in the numerator cancels with the denominator, leaving:

(20*19*18)/(3*2*1) = 1140

Again, this will only make sense if you've studied these concepts.

This number has to be reduced by 18 because the question restricts us from including any 3 immediately adjacent students.

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Re: From a class of 20 students whose names are listed in alphabetical. [#permalink]
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safety16644 wrote:
Yes, however I dont understand how they computed that number and arrived to 1140, hoping for a little bit more detail on it. 

20 C3 =1140 doesnt make sense to me, how did they arrive to 1140?

­When choosing groups of 3 out of 20 students, we can choose them in the following way.

There are 20 choices for the first student.

Then, we have 19 choices for the second student.

Finally, we have 18 choices left for the third student.

So, for any of the 20 students chosen first, we have 19 different ways to choose the second and then 18 different ways to choose the third.

Accordingly, we can multiply to find the number of possible groups. 20 × 19 × 18.

However, we aren't done, because that resulting number will include duplicate groups. For instance, if there are 3 students, A, B, and C, our total will include ABC, BCA, CBA, etc.

So, we divide by 3! to eliminate the duplicates.

Thus, we have 20 × 19 × 18/3! = 1,140
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From a class of 20 students whose names are listed in alphabetical. [#permalink]
From a class of 20 students whose names are listed in alphabetical order, a teacher will choose one group of 3 students to represent the class in a student congress. If the teacher will not choose a group of 3 students whose names are in 3 consecutive positions on the list, how many different groups of 3 students could be chosen by the teacher?

(A) 1120
(B) 1122
(C) 1135
(D) 1137
(E) 1140


Nice question. Doing this indirectly may be helpful. First, let's imagine the 20 students are lettered: a, b, c, d, e... (up to the 20th student)

Total selections possible if we could choose any 3 letters: 20C3 which is 1140

Selections possible in which 3 letters are consecutive: Out of the 20 letters, only 18 letters can have 2 consecutive letters attached. For example, there is no letter you can put after the 20th since there is no 21st or 22nd letter available.

Selections possible in which 3 letters are NOT consecutive: 1140 - 18 = 1122

(B) is your answer.­
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Re: From a class of 20 students whose names are listed in alphabetical. [#permalink]
Hello MartyMurray could you help me understand how 1140 gets reduced to 1122?
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Re: From a class of 20 students whose names are listed in alphabetical. [#permalink]
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ashdank94 wrote:
From a class of 20 students whose names are listed in alphabetical order, a teacher will choose one group of 3 students to represent the class in a student congress. If the teacher will not choose a group of 3 students whose names are in 3 consecutive positions on the list, how many different groups of 3 students could be chosen by the teacher?

(A) 1120
(B) 1122
(C) 1135
(D) 1137
(E) 1140

Hello MartyMurray could you help me understand how 1140 gets reduced to 1122?

­The TOTAL number of ways to select any 3 students from 20 is 20C3 = 20!/(17!3!) = 1,140.

The number of 3-student groups whose names ARE in 3 consecutive positions on the list is 18 (assuming the students are listed in alphabetical order):

(1, 2, 3)
(2, 3, 4)
(3, 4, 5)
...
(18, 19, 20)

When we subtract the number of 3-student groups whose names ARE in 3 consecutive positions from the TOTAL number of ways to select any 3 students from 20, we get the number of 3-student groups whose names are NOT in 3 consecutive positions.

Therefore, the answer is 1,140 - 18 = 1,122.

Answer: B.­
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Re: From a class of 20 students whose names are listed in alphabetical. [#permalink]
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