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Re: A certain club has 10 members, including Harry. One of the [#permalink]
Dear IanStewart GMATGuruNY VeritasKarishma,

Combinatorics can't be used to solve this question because we deal with RANKS/ORDER (i.e. secretary and treasurer), right?

I hesitate because I saw similar question in which Bunuel used Combinatorics to solve the problem.(https://gmatclub.com/forum/a-certain-cl ... 27730.html)
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Re: A certain club has 10 members, including Harry. One of the [#permalink]
Dear IanStewart,

I meant using Combination (choosing a subgroup from a larger group) formula as quoted below:
Bunuel wrote:
One can also do: \(P=\frac{C^1_1*C^2_7}{C^3_8}=\frac{3}{8}\), where \(C^1_1\) is # of ways to select Kim, \(C^2_7\) is # of ways to select any 2 students out of 7 left and \(C^3_8\) is total # of ways to select 3 students from 8;

Or: you can find the probability that among 3 students selected to complete the tasks there won't be Kim and subtract it from 1:
\(P=1-\frac{C^3_7}{C^3_8}=\frac{3}{8}\);

I tried and failed to apply the above method to this question :(

BTW, I know that permutation works for this problem. But not sure about the combination one.
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Re: A certain club has 10 members, including Harry. One of the [#permalink]
Dear IanStewart
IanStewart wrote:
1.: 10C2 ways to pick the Secretary and Treasurer, and 9C1*1C1 to make one of those selections Harry, so 9C1/10C2 = 1/5 is the answer

2.: 9C2 ways to pick the Secretary and Treasurer if Harry will not be selected, and 10C2 selections possible in total, so 1 - (9C2/10C2) = 1 - 4/5 = 1/5 is the answer

I don't understand why we can completely ignore "president" selection?
How can we calculate as if there were no "president" selection at all?
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Re: A certain club has 10 members, including Harry. One of the [#permalink]
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varotkorn wrote:
I don't understand why we can completely ignore "president" selection?
How can we calculate as if there were no "president" selection at all?


The fact that they're picking a President doesn't matter, since we don't care if Harry is President. You can just imagine picking the Secretary and Treasurer first. Then it doesn't matter what they do with the remaining eight people - whether they give them all positions like Chairperson, President, Vice-President, etc, or give none of them positions. Harry will be just as likely either way to be Secretary or Treasurer.

This problem is identical, mathematically, to this one: you have 10 people, including Harry. They all line up. What is the probability Harry is 2nd or 3rd in line? Whether you call the first person in line "President" or don't call that person anything, the answer is the same either way.
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Re: A certain club has 10 members, including Harry. One of the [#permalink]
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Some of you are overthinking this explanation!

As have others earlier on this thread, I recommend that you keep it simple. Don't count possibilities and make the math complicated--go straight to the probabilities, since that's the question being asked.

Concept: "or" probabilities can be added, so long as you remember to subtract "both" in the cases where that's an option.

Explanation: 1/10 + 1/10 = 2/10 = 1/5

That's it.

-Brian
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Re: A certain club has 10 members, including Harry. One of the [#permalink]
Bunuel, GMATNinja, NYCgirl15, why we don't multiply by 7/8 to find

NOT President but Secretary = 9/10*1/9=1/10? (My understanding is that when selected as secretary, he cannot be selected as President or Treasurer. Negative prob for president is 9/10 and negative probability for treasurer is 7/8. Therefore, NOT President but Secretary = 9/10*1/9=1/10*7/8) Where I am getting wrong with my analysis?


President= 1/10
Not President= 1-1/10=9/10

Secretary= 1/9
Not Secretary= 1-1/9=8/9

Treasurer= 1/8
Not Treasurer= 1-1/8= 7/8

Thus,
1. NOT President but Secretary = 9/10*1/9=1/10
2. NOT President and NOT Secretary but Treasurer= 9/10*8/9*1/8=1/10
3. Either Secretary or Treasurer= 1/10+1/10=2/10=1/5
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Re: A certain club has 10 members, including Harry. One of the [#permalink]
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tkorzhan1995 wrote:
Bunuel, GMATNinja, NYCgirl15, why we don't multiply by 7/8 to find

NOT President but Secretary = 9/10*1/9=1/10? (My understanding is that when selected as secretary, he cannot be selected as President or Treasurer. Negative prob for president is 9/10 and negative probability for treasurer is 7/8. Therefore, NOT President but Secretary = 9/10*1/9=1/10*7/8) Where I am getting wrong with my analysis?


President= 1/10
Not President= 1-1/10=9/10

Secretary= 1/9
Not Secretary= 1-1/9=8/9

Treasurer= 1/8
Not Treasurer= 1-1/8= 7/8

Thus,
1. NOT President but Secretary = 9/10*1/9=1/10
2. NOT President and NOT Secretary but Treasurer= 9/10*8/9*1/8=1/10
3. Either Secretary or Treasurer= 1/10+1/10=2/10=1/5

By using "1/9" in your calculation of "NOT President but Secretary," you are picking Harry as the Secretary. Because you've already selected him as Secretary, he is no longer part of the remaining 8 people.

In other words, none of the remaining 8 people are Harry, so ANY of those 8 can be chosen as Treasurer (not just 7 of the 8).

I hope that helps a bit!
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Re: A certain club has 10 members, including Harry. One of the [#permalink]
IanStewart
could you or someone help explain your approach on 1- P(not harry). I am not fully how 9C2 would show the various outcomes of picking not harry for treasurer or secretary. Are we just assuming since 1 of the 10 is picked for president (including harry as a possibility), then you subtract that 1 person from the 10, giving you 9 remnatns that could be treasurer or secretary? (what if Harry was among the 9)? please help thanks!

while the first approach somewhat made a better case for me that 1C1 for harry as a secretary or treasurer, since harry has been picked, the remaining 9 people could be chosen for the whichever position harry was not picked for ie., so you have a 9C1*1C1 way of arranging the desired outcome.
I get the concept of subtracting the chances that harry is not either of the position.
I understand the very simplified explanation but I thought the combinatoric formula helped me understand some underlying understanding of the stem.
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Re: A certain club has 10 members, including Harry. One of the [#permalink]
gmatophobia

Can you please help me understand the mistake I am making in the following solution

Selected as secretary
P(Secretary)*P(treasury)*P(President) = (1C1 * 9C1 * 8C1) / 10C3 = 6/10

Selected as treasurer
P(treasury)*P(Secretary)*P(President) = (1C1 * 9C1 * 8C1) / 10C3 = 6/10

Thus
6/10 + 6/10 = 12/10 = 1.2 ...obviously wrong since probability cannot be >1
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A certain club has 10 members, including Harry. One of the [#permalink]
Rickooreo wrote:
gmatophobia

Can you please help me understand the mistake I am making in the following solution

Selected as secretary
P(Secretary)*P(treasury)*P(President) = (1C1 * 9C1 * 8C1) / 10C3 = 6/10

Selected as treasurer
P(treasury)*P(Secretary)*P(President) = (1C1 * 9C1 * 8C1) / 10C3 = 6/10

Thus
6/10 + 6/10 = 12/10 = 1.2 ...obviously wrong since probability cannot be >1


Rickooreo - I couldn't follow your approach. I understood that you're taking the combinations.

There are two ways to go about this -

Using Probability

Select three members & Assign them one of the titles

Probability of Harry getting selected for the one of the three positions = 3/10

Once selected, Harry there are two favorable roles. Probability of Harry Getting treasurer or secretary = 2/3

Net probability = \(\frac{2}{3} * \frac{3}{10} = \frac{1}{5}\)

Using Combinations

If you want to approach this questions P&C -

Total number of cases = Select 3 people from the 10 members in the club and give them one of the three available titles.

= 10C3 * 3!

Favorable cases -

Because we need Harry to be a part of the committee, we need two more people from 9 available members of the club. So we can select 2 people out of 9 available members in 9C2 ways.

Once selected, we can assign the titles among these three people in 3! ways, however we need to discard two ways in which Harry is the President. So the number of favorable arrangement in each group is 4.

Net Probability = \(\frac{9C2 * 4 }{ 10C3 * 6}\) = \(\frac{1}{5}\)
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Re: A certain club has 10 members, including Harry. One of the [#permalink]
I am not strong in probability formula. However, I have an option to look at the problem from a plain arithmetic point of view. First of all the president will be chosen. If Harry hasn’t been chosen as the president, then he has the chance to be eihter secretary or the treasurer. Given that, once someone other than Harry has already been chosen as the president, now 9 members are remaining from whom first the secretary will be chosen. Similarly when the secretary has already been chosen, then the treasurer will be chosen from the remaining 8 members. Therefore, Harry's chance to be the secretary is 1/9, thus his chance of being the treasurer is 1/8. From this logic, the average of 1/9 and 1/8 is 17/72.
I would really appreciate your help.
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Re: A certain club has 10 members, including Harry. One of the [#permalink]
To calculate the probability that Harry will be either the member chosen to be the secretary or the member chosen to be the treasurer, we need to consider two scenarios:

Scenario 1: Harry is chosen as the secretary.
In this case, we have 9 remaining members to choose from for the president. After selecting the president, we have 8 remaining members to choose from for the treasurer. So the probability of Harry being chosen as the secretary is 1/10 (since there are 10 members initially) multiplied by 9/9 (since any of the remaining 9 members can be chosen as the president) multiplied by 8/8 (since any of the remaining 8 members can be chosen as the treasurer). This gives us a probability of 1/10.

Scenario 2: Harry is chosen as the treasurer.
Similar to Scenario 1, we have 9 remaining members to choose from for the president. After selecting the president, we have 8 remaining members to choose from for the secretary. So the probability of Harry being chosen as the treasurer is 1/10 (since there are 10 members initially) multiplied by 9/9 (since any of the remaining 9 members can be chosen as the president) multiplied by 8/8 (since any of the remaining 8 members can be chosen as the secretary). This also gives us a probability of 1/10.

Since we are interested in either of these two scenarios happening, we can add their probabilities together:
1/10 + 1/10 = 1/5

Therefore, the probability that Harry will be either the member chosen to be the secretary or the member chosen to be the treasurer is (E) 1/5.
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Re: A certain club has 10 members, including Harry. One of the [#permalink]
Translating this we're finding:

P(NOT PRESIDENT) * P(YES SECRETARY) + P(NOT PRESIDENT) * P(NO SECRETARY) * P(YES TREASURER)

If he becomes secretary in the first scenario, then we can omit the changes of him becoming treasurer since the order the choose is president -> secretary -> treasurer.

9/10 * 1/9 + 9/10 * 8/9 * 1/8 = 1/10 + 1/10 = 2/10 = 1/5
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Re: A certain club has 10 members, including Harry. One of the [#permalink]
Hi there! Can anyone help me understand what is the total outcome? Is it 720 or 120? I see it as 3C10 hence 120. I see in the discussion the majority see it as 10*9*8 (720), but in some cases it is also seen as 120...
I am confused.... I was using the approach (favourable / all)
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Re: A certain club has 10 members, including Harry. One of the [#permalink]
sagarsir wrote:
E.

[ 9C2 * ( 3! - 2!) ] / 10C3
= 1/5

­
Why did you multiply 9C2 with 3! - 2!?
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Re: A certain club has 10 members, including Harry. One of the [#permalink]
ATQ... All the members have an equal chance of getting selected for each post because it is a random draw, AND nowhere is it written that the selection are going to happen in the "exact order" given. So basically the chances for harry to be picked out for either post is 1/10 and there are 2 such posts that we're asked to get the probability of

so 1/10 + 1/10 = 2/10 = 1/5
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