From a class of 20 students whose names are listed in alphabetical.

Question

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?

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

Bunuel · Accepted Answer

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.

gmatophobia · Answer

Let's assume that the names of 20 students appear on the list as shown below.  S_n  represents the name of the  n^{	ext{th}}  student

S_1 \quad S_2 \quad S_3 \quad S_4 \quad S_5 \quad S_6 \quad S_7 .................... \quad S_{17} \quad S_{18} \quad S_{19} \quad S_{20}

If there were no restrictions, the teacher could have chosen the three students in  ^{20}C_3  ways. However, the teacher cannot choose 3 consecutive positions on the list. Let's try to partition the list such that we can select three consecutive students between the partition.

| | → denotes the partition

| S_1 \quad S_2 \quad S_3| \quad S_4 \quad S_5 \quad S_6 \quad S_7 .................... \quad S_{17} \quad S_{18} \quad S_{19} \quad S_{20}

S_1 \quad |S_2 \quad S_3 \quad S_4| \quad S_5 \quad S_6 \quad S_7 .................... \quad S_{17} \quad S_{18} \quad S_{19} \quad S_{20}

S_1 \quad S_2 \quad |S_3 \quad S_4 \quad S_5 | \quad S_6 \quad S_7 .................... \quad S_{17} \quad S_{18} \quad S_{19} \quad S_{20} 
.
.
 S_1 \quad S_2 \quad S_3 \quad S_4 \quad S_5 \quad S_6 \quad S_7 .................... \quad | S_{17} \quad S_{18} \quad S_{19} | \quad S_{20}

S_1 \quad S_2 \quad S_3 \quad S_4 \quad S_5 \quad S_6 \quad S_7 .................... \quad S_{17} \quad | S_{18} \quad S_{19} \quad S_{20} |

There are 18 ways the students can be partitioned such that three consecutive students appear; we need to subtract this number from  ^{20}C_3 .

Note : We don't have to perform the partition as represented above. I have done so to represent the ways the partition will occur. We can simply count the start or the end positions of the partition to determine the number. For example, the first partition ends after  S_3 , and the last partition ends after  S_{20}

Number of partitions =  (20 - 3) + 1 = 18

Total possible ways =  ^{20}C_3 - 18 = 1140 - 18 = 1122

Option B

youcouldsailaway · Answer

The total number of ways through which 3 members can be chosen is  20 C 3  = 1140.

We can immediately remove option E from consideration since 1140 is the comprehensive list of combinations, which we wish to further modify.

Now, if we assign a letter to each individual then we will have individuals from A to T in our list.

The teacher will not choose individuals who appear consecutively in the list. Thus, the following combinations are out:

ABC
BCD
CDE
DEF
EFG
...
...
...
...
RST

From A (1st letter of the alphabet) to R (18th letter of the alphabet), that's 18 total such combinations. To elaborate, there is no person before A so the combinations begin with A. Since T is the 20th person in the list, R S T is the only possible last combination - since we cannot for instance take S T U, as there is no individual U in our list.

Thus, subtracting 18 from 1140, we get 1122. Thus, B is the answer.

safety16644 · Answer

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?

Regor60 · Answer

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.

Posted from my mobile device

MartyMurray · Answer

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

ashdank94 · Answer

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

tuyaras · Answer

No need to count manually, you just write and you will see the pattern

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

As you can see, you can get number of the case by looking at the first number in each case.
So we will get total 18 cases to deduct.

CapnSal · Answer

Bunuel , is this a correct logic to find the count of consecutive entries? Any ordered list of length N, number of ways to pick K consecutive entries is given by: N - K +1

Here, N = 20, K = 3. So 20-3+1 = 18.

Otherwise what is an efficient way without manually writing all the cases?

sethiayush · Answer

One can select 3 consecutive positions on the list by considering,

3 kids as 1 unit
This leaves you with 17+1 = 18 units to be selected where 3 consecutive positions occur together

This selection can happen in 18C1 ways

Therefore, your answer becomes 20C3 - 18C1

lawyeroffduty · Answer

hey, if you have found the answer can you please share it here

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=86743 --- Harsha GMAT-Club-Forum-torpm0yx.png

Vaishbab · Answer

I understood the question eventually but was it just me who was thrown off by "a group of 3 students..."? I thought the question was talking about a singular group of students in consecutive positions. If that was the case, the answer would be 1139 and thankfully, that option wasn't available.

Shouldn't it be "any group of 3 students..."?

Shouldn't it be " any  group of 3 students..."?

Bunuel · Answer

The wording is already correct. “A group of 3 students” means any possible group of 3, not one fixed group. If the question meant only one specific consecutive group, it would say “the group of 3 consecutive students,” not “a group of 3 students.”

From a class of 20 students whose names are listed in alphabetical.

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From a class of 20 students whose names are listed in alphabetical.

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