M37-20

Question

Harry is planning a journey to Hogwarts. He can go alone or with any number of his 7 friends: Ron, Hermione, Hagrid, Luna, Neville, Fred and George. If Ron and Hermione refuse to go together, how many groups are possible for the journey ?

A. $88$
B. $95$
C. $96$
D. $1,560$
E. $3,600$

Show Answer

Official Answer

C

A.  88 
B.  95 
C.  96 
D.  1,560 
E.  3,600

Bunuel · Answer

Official Solution:

Harry is planning a journey to Hogwarts. He can go alone or with any number of his 7 friends: Ron, Hermione, Hagrid, Luna, Neville, Fred and George. If Ron and Hermione refuse to go together, how many groups are possible for the journey ?

A. $88$
B. $95$
C. $96$
D. $1,560$
E. $3,600$

Show Answer

Official Answer

C

A.  88 
B.  95 
C.  96 
D.  1,560 
E.  3,600

How many groups are possible if we did not have the restriction? Without the restriction the total number of groups possible is  2^7 : each of Harry's 7 friends can either join Harry or not. 
 
How many groups are possible with Ron and Hermione in them? If Ron and Hermione are in the group, then each of the 5 remaining friends can either join or not, so the number of groups with Ron and Hermione is  2^5 .
 
So, there are  Total-Restriction=2^7-2^5=2^5(2^2-1)=96  groups possible.

Answer: C

einstein801 · Answer

Question stem says ron and hermione refuse to go together. Answer stem includes ron and hermione going together. Please revise question/answer thank you.

Bunuel · Answer

It seems you did not read the solution carefully. The question indeed asks to find the number of groups without Ron and Hermione together. We find this by subtracting the number of groups  with  them (restriction condition) from the  total  number of groups, thus getting the number of groups  without  them. This is clearly given in the solution: "Total - Restriction = 2^7 - 2^5 = 2^5(2^2-1) = 96".

tukaisg · Answer

To determine the number of possible groups for the journey, we need to consider the constraints and available choices:
Total combinations: Harry can travel alone or with any combination of his 7 friends. Without any constraints, there are $2^7 = 128$ possible combinations, including the option of traveling alone.
Constraint (Ron and Hermione can't go together): We need to exclude the combinations where both Ron and Hermione are part of the group. If Ron and Hermione are together, they can be accompanied by any combination of the remaining 5 friends, which gives $2^5 = 32$ combinations.
Valid combinations: Subtract the invalid combinations from the total: $128 - 32 = 96$.
Thus, there are 96 possible groups for Harry's journey to Hogwarts under the given constraints.

The calculation of $2^7$ comes from considering the number of ways to choose subsets of Harry's 7 friends. Each friend can either be included in the group or not, which gives us two choices (include or exclude) per friend. Thus, for 7 friends, the total number of possible combinations is $2^7$.
This includes all possible groups, ranging from no friends (Harry traveling alone) to all 7 friends traveling with him.

To elaborate on the constraint that Ron and Hermione can't travel together:
Excluding Ron and Hermione Together: When considering the combinations where both Ron and Hermione are traveling together, we start by fixing them in the group. Once they are chosen, the remaining decision involves the other 5 friends (Hagrid, Luna, Neville, Fred, and George).
Combinations with Remaining Friends: For each of these 5 friends, there are 2 choices: either they join the group or they don't. This gives us $2^5$ combinations because each friend has two possibilities. Therefore, with Ron and Hermione already fixed, there are $32$ different ways to form a group with any subset of the remaining 5 friends.
By excluding these 32 combinations from the total, we ensure that no group includes both Ron and Hermione together, thus satisfying the constraint.

Nipunh · Answer

Harry can choose groups of friends in 3 main scenarios:
1) Ron is included, but not Hermione.
2) Hermione is included, but not Ron.
3) Neither Ron nor Hermione is included.
We'll count the number of valid groups for each scenario.

Case 1: Ron is included, but not Hermione
If Ron is included, then Hermione cannot be part of the group. The remaining friends to consider are: {Hagrid,Luna,Neville,Fred,George}(5 people)
Harry can choose any subset of these 5 friends (including no one), giving us 
5C0+5C1+5C2+5C3+5C4+5C5= 1+5+10+10+5+1=32 groups (we'll use reasoning rather than formulas: for each friend, Harry either includes them or doesn't, leading to all possible subsets).

Case 2: Hermione is included, but not Ron
If Hermione is included, then Ron cannot be part of the group. The remaining friends to consider are:

Just as in Case 1, Harry can choose any subset of these 5 friends. Again, there are:
{Hagrid,Luna,Neville,Fred,George}(5 people)
Harry can choose any subset of these 5 friends (including no one), giving us 
5C0+5C1+5C2+5C3+5C4+5C5= 1+5+10+10+5+1=32 groups

4. Case 3: Neither Ron nor Hermione is included
If neither Ron nor Hermione is included, the remaining friends are
{Hagrid,Luna,Neville,Fred,George}(5 people). Once again, Harry can choose any subset of these 5 friends. This gives: 32 valid groups.
5. Total valid groups
Add up the groups from all three cases:
32+32+32=96

Javmex · Answer

Hi  Bunuel

Does the word Any includes 0 ?

Thanks in advance!

Bunuel · Answer

Yes, and with both 2^7 and 2^5, you would include the case when none of the friends are selected, resulting in 0 friends. You see, 2^7 comes from each of the 7 friends having two options: either to join Harry or not. Hence, we get 2 * 2 * 2 * 2 * 2 * 2 * 2 = 2^7. One of these cases among the 2^7 possibilities is the scenario where none of the friends joins Harry.

Hope it's clear.

SudsMeister · Answer

I did it in a much lengthier way as the optimal method didn't strike me.
It goes like this:

Harry goes alone = 1 way.

Harry goes with one friend = Selecting 1 friend out of 7  = 7C1 =   7 ways.

Harry goes with 2 friends = Selecting 2 friends out of 7  -  Selecting 1 way in which Ron and Hermione go together  = 7C2 - 1 = \frac{7*6}{2*1} - 1 = 21 - 1 =  20 ways.

Harry goes with 3 friends = Selecting 3 friends out of 7  -  Selecting 1 friend out of 5 as Ron and Hermione go together  = 7C3 - 5C1 =\frac{7*6*5}{3*2*1} - 5 = 35 - 5 =  30 ways.

Harry goes with 4 friends = Selecting 4 friends out of 7  -  Selecting 2 friends out of 5 as Ron and Hermione go together  = 7C4 - 5C2 =\frac{7*6*5}{3*2*1} - \frac{5*4}{2*1} = 35 - 10 =  25 ways.

Harry goes with 5 friends = Selecting 5 friends out of 7  -  Selecting 3 friends out of 5 as Ron and Hermione go together  = 7C5 - 5C3 = \frac{7*6}{2*1} - \frac{5*4}{2*1} = 21 - 10 =  11 ways.

Harry goes with 6 friends = Selecting 6 friends out of 7  -  Selecting 4 friends out of 5 as Ron and Hermione go together  = 7C6 - 5C4 = 7 - 5 =  2 ways.

Harry goes with 7 friends = Not possible as Ron and Hermione cannot travel together = 0 way.

Total ways = 1 + 7 + 20 + 30 + 25 + 11 + 2 + 0 = 96 ways.

Answer C.

Rebaz · Answer

I did not quite understand the solution.  Bunuel

when you say each of Harry's 7 friends can either join Harry or not, do you mean that each group exists of 2 persons?

(Haryy+Ron) and (Haryy+Hermoine) and (Haryy+Hagrid) and (Haryy+Luna) and (Haryy+Neville) and (Haryy+Fred) and (Haryy+George) and hence 2^7 ???

Thanks in advance!

Bunuel · Answer

No. We are not forming only 2-person groups. “Each of Harry’s 7 friends can either join him or not” means we’re counting all possible combinations of his friends, from no one going with him to all 7 joining. That gives 2 choices (in or out) for each of the 7 friends, so the total is 2^7 = 128 possible groups (including the empty group where Harry goes alone). For example, one of the combinations is where all 7 friends go with Harry. That group is included in the 2^7. Another is where only Ron joins, or only Luna, or where no one goes at all. All of these cases are covered. Here are more solutions: https://gmatclub.com/forum/harry-is-pla ... 65289.html

sb995 · Answer

I like the solution - it’s helpful. This is a good question.

M37-20

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