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Expert plz help with combinatorics using SLOT METHOD

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Expert plz help with combinatorics using SLOT METHOD

by Redhorsep » Thu Aug 25, 2011 12:04 pm
Hi,

I tried to solve a lot of combinatoric problems using the Slot Method taught by Ron in the study hall recording. But it seems hard to use it for problems that have certain restrictions/designated persons included, please show me how you would do these problems with Slot Method, or better alternative approach:

# 1: A basketball-ball coach will pick a five-player team from a group of 7 players. If Mary and Patricia are among the 7 players the coach can pick from, how many different teams that include Mary but not Patricia can the coach form?





# 2: Jane and Keisha are among the six executives Company XYZ can pick from to send a four-executive delegation to a business conference. How many different four-executive groups can Company XYZ send to the conference if both Jane and Keisha must be included?



# 3: John and Mike are among the runners of a five-people foot-race, how many possible finishes are there if John must arrive ahead of Mike?


Also, which way do you think it's better? the Slot Method or the Anagram method (MGMAT)? I found the first one easier to work with and quicker. Thanks a lot!
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by BenchPrepGURU » Thu Aug 25, 2011 12:57 pm
I find both the slot method and the anagram method unreliable for harder problems. This is generally what happens when you learn a trick, without the reasoning behind the trick, to solve a certain kind of problem. It works for a while, but when you encounter harder problems it breaks down. The real trick is to understand the concept.

The two main concepts in these kind of problems are permutations and combinations. Both involve picking r things at random from a set of n things where n is larger than r.

A permutation is where you pick one thing at a time and assign it a place in some order
A combination is where you pick a group of things all together with no regard to order

The slot method is custom built for permutations. Combinations can be found by calculating a permutation and then dividing by the number of ways that permutation can be reordered.

For permutations, I'll use the notation P(n, r); for combinations, C(n, r)
P(n, r) = n!/(n-r)!
C(n, r) = n!/[(n-r)!r!] = P(n, r)/r!
Note that 0! = 1 by definition

# 1: A basketball-ball coach will pick a five-player team from a group of 7 players. If Mary and Patricia are among the 7 players the coach can pick from, how many different teams that include Mary but not Patricia can the coach form?

Here, because none of the players is assigned a position or any place in an order, we'll use combinations. We can break the players into 3 groups:

Mary (pick 1) = C(1, 1) = 1
Patricia (pick 0) = C(1, 0) = 1
Other (pick 4) = C(5, 4) = 5!/(4!*1!) = 5!/4! = 5

Number of teams = 1*1*5 = 5

A simpler way to think about this: you've picked Mary. There are six people left, but you can't pick Patricia. Instead of thinking about how many way you can pick the remaining 4 players, think about the number of ways you can pick the one player in addition to Patricia who will not play. 5 players to pick from - 5 ways.

# 2: Jane and Keisha are among the six executives Company XYZ can pick from to send a four-executive delegation to a business conference. How many different four-executive groups can Company XYZ send to the conference if both Jane and Keisha must be included?

Again, this is a combination - delegation is just a subset of a larger group with no regard to order. We break up the six executives as follows

Jane and Keisha (pick 2) = C(2, 2) = 1
The other 4 Execs (pick 2) = C(4, 2) = 6

Number of delegations = 1*6 = 6

# 3: John and Mike are among the runners of a five-people foot-race, how many possible finishes are there if John must arrive ahead of Mike?

Now order matters. This would be an easy problem if John arrived directly ahead of Mike, but the way it's written Mike can arrive anywhere after John. There are some slick ways to solve this, but they don't really make sense in a time-pressure situation, so I'm just going to brute force it (I find that many of my students are convinced that there is a trick to every GMAT problem and the people who score high know all the tricks. There may be a trick to every problem, but often brute force will get the problem done in under 2 minutes, when thinking of the slick trick will take at least 3 minutes - I used brute force a lot and I scored a 770)

John 1st - now the other four can finish in any order = 1*4*3*2*1 = 24
or
John 2nd - now Mike can't be first so = 3*1*3*2*1 = 18
or
John 3rd - now Mike can't be first or second = 3*2*1*2*1 = 12
or
John 4th - now Mike has to be last = 3*2*1*1*1 = 6

John can't finish 5th. Since these are mutually exclusive "or" scenarios, we add
24 + 18 + 12 + 6 = 60

In sum, there's no ultimate method. Be flexible - this means not being afraid to grind out a tedious calculation

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by GMATGuruNY » Thu Aug 25, 2011 8:28 pm
# 1: A basketball-ball coach will pick a five-player team from a group of 7 players. If Mary and Patricia are among the 7 players the coach can pick from, how many different teams that include Mary but not Patricia can the coach form?
Question rephrased: How many ways can we choose a combination of 4 players to be put together with Mary?
Since Patricia cannot be included, we need to choose a combination of 4 from the 5 other players.
Using the slot method:
(5*4*3*2)/(4*3*2*1) = 5.
# 2: Jane and Keisha are among the six executives Company XYZ can pick from to send a four-executive delegation to a business conference. How many different four-executive groups can Company XYZ send to the conference if both Jane and Keisha must be included?
Question rephrased: How many ways can we choose a combination of two executives to be put together with Jane and Keisha?
We need to choose a combination of 2 from the 4 other executives.
Using the slot method:
(4*3)/(2*1) = 6.
# 3: John and Mike are among the runners of a five-people foot-race, how many possible finishes are there if John must arrive ahead of Mike?
Number of ways to arrange 5 elements = 5*4*3*2*1 = 120.
In half of these arrangements, John will be ahead of Mike; in the other half, Mike will be ahead of John.
Number of arrangements in which John is ahead of Mike = (1/2)*120 = 60.

For a more detailed explanation of the slot method, check my post here:

https://www.beatthegmat.com/pls-clear-my ... 85743.html
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
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by Redhorsep » Sat Aug 27, 2011 12:06 pm
GMATGuruNY wrote:
# 1: A basketball-ball coach will pick a five-player team from a group of 7 players. If Mary and Patricia are among the 7 players the coach can pick from, how many different teams that include Mary but not Patricia can the coach form?
Question rephrased: How many ways can we choose a combination of 4 players to be put together with Mary?
Since Patricia cannot be included, we need to choose a combination of 4 from the 5 other players.
Using the slot method:
(5*4*3*2)/(4*3*2*1) = 5.
# 2: Jane and Keisha are among the six executives Company XYZ can pick from to send a four-executive delegation to a business conference. How many different four-executive groups can Company XYZ send to the conference if both Jane and Keisha must be included?
Question rephrased: How many ways can we choose a combination of two executives to be put together with Jane and Keisha?
We need to choose a combination of 2 from the 4 other executives.
Using the slot method:
(4*3)/(2*1) = 6.
# 3: John and Mike are among the runners of a five-people foot-race, how many possible finishes are there if John must arrive ahead of Mike?
Number of ways to arrange 5 elements = 5*4*3*2*1 = 120.
In half of these arrangements, John will be ahead of Mike; in the other half, Mike will be ahead of John.
Number of arrangements in which John is ahead of Mike = (1/2)*120 = 60.

For a more detailed explanation of the slot method, check my post here:

https://www.beatthegmat.com/pls-clear-my ... 85743.html



Hi Mitch, I wonder how you would solve this problem using the SLOT METHOD. I believe this is a special type of permutation problem and I remember for circular permutation it should be calculated as (n-1)!

I tried it myself and got the answer to match the official answer but I want to see if my thought process was correct, can you show me yours?

"Juan's sound system has a circle-shaped automatic CD-changer with six slots for CDs arranged on a circle. If Juan has six CDs, in how many different arrangements can his six CDs be placed on the CD-changer if the blues CD must be put into slot number 1 and the rock CD into one of the slots next to slot number 1?"

Thanks again!
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by GMATGuruNY » Sat Aug 27, 2011 12:57 pm
Redhorsep wrote: "Juan's sound system has a circle-shaped automatic CD-changer with six slots for CDs arranged on a circle. If Juan has six CDs, in how many different arrangements can his six CDs be placed on the CD-changer if the blues CD must be put into slot number 1 and the rock CD into one of the slots next to slot number 1?"
Since the blues CD must be placed in slot 1, the number of options for the blues CD = 1.
Since the rock CD must be in one of the 2 slots adjacent to slot 1, the number of options for the rock CD = 2.
The number of ways to arrange the 4 remaining songs in the remaining 4 slots = 4*3*2*1.
Multiplying the numbers above, we get:
Total number of arrangements = 1*2*4*3*2*1 = 48.

Check here for another question about circular arrangements:

https://www.beatthegmat.com/seating-arra ... 85488.html
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

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by saketk » Sun Aug 28, 2011 10:38 pm
Redhorsep wrote:Hi,

I tried to solve a lot of combinatoric problems using the Slot Method taught by Ron in the study hall recording. But it seems hard to use it for problems that have certain restrictions/designated persons included, please show me how you would do these problems with Slot Method, or better alternative approach:


Also, which way do you think it's better? the Slot Method or the Anagram method (MGMAT)? I found the first one easier to work with and quicker. Thanks a lot!
Rather than using a pre-determined technique -- use basics...

Redhorsep wrote:# 1: A basketball-ball coach will pick a five-player team from a group of 7 players. If Mary and Patricia are among the 7 players the coach can pick from, how many different teams that include Mary but not Patricia can the coach form?
This questions will be easy if you start something like this--
Mary already in the team. Since Mary & Patricia cannot be together, Patricia is out of selection process. People Left for selection =5. out of 5 we need 4 more (Patricia is already in) -- 5C4 = 5 ways
Redhorsep wrote:# 2: Jane and Keisha are among the six executives Company XYZ can pick from to send a four-executive delegation to a business conference. How many different four-executive groups can Company XYZ send to the conference if both Jane and Keisha must be included?
If 'J' and 'K' are already in, Question reduces to -- how many ways 2 Executive can be selected from a team of 4 executive -- ANSWER -- 4C2= 6 ways
Redhorsep wrote:# 3: John and Mike are among the runners of a five-people foot-race, how many possible finishes are there if John must arrive ahead of Mike?
John needs to be ahead of Mike-- this means John can take any place except the last place.

There will be 4 cases-
John is 1st (the next Usian Bolt :) )
The other 4 can be arranged in 4! ways
John is 2nd --this way Mike can choose from 3rd,4th & 5th place only
The total number of ways for Mike-- 3 and for other 3 = 3!
Likewise we will have 2 more case
3rd place -- 2*3!
4th place -- 1*3!
Total = 4! + 3! (1+2+3) = 24+36 = 60 ways

Don't stick to a particular method for PnC or shall I say for Math as a whole-- In most of the cases a math question can be solved with different methods!
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by saketk » Sun Aug 28, 2011 11:43 pm
For cicular arrangement--
(a)If clockwise and anti clock-wise orders are different, then total number of circular-permutations is given by (n-1)!

(b)If clock-wise and anti-clock-wise orders are taken as not different, then total number of circular-permutations is given by (n-1)!/2!
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by garry9839 » Tue Nov 26, 2013 11:09 pm
I really appreciate if you could help me to solve the questions by slot method


below are some questions

1-There are 3 states and three students representing each state .in how many ways 5 students can be chosen such that at least one student is chosen from each state?
2-How would you divide 5 distinct objects in a groups of 2, 2 and 1 ?

3-8 man has to be chosen from the 12 man for rowing competition such that 3 of them has always to sit at one side of boat and one at other side .in how many ways we can do this if 4 people has to sit each side?

4-Acrewof an eight oar boat has to be chosen out of 11men five of whom can row on stroke side only, four
on the bow side only, and the remaining two on either side.How many different selections can be made?

5-(a) Howmany divisors are there of the number x = 21600. Find also the sumof these divisors.
(b) In howmanyways the number 7056 can be resolved as a product of 2 factors.
(c) Find the number of ways in which the number 300300 can be split into 2 factors which are
relatively prime.
Number of divisors of the form 4n + 2 ( n>= 0 )of the integer 240 is

6-If r, s, t are prime numbers and p, q are the positive integers such that their LCMof p, q is is r2t4s2, then
the numbers of ordered pair of (p, q) is



sorry for my poor English

Regards

Gaurav



GMATGuruNY wrote:
Redhorsep wrote: "Juan's sound system has a circle-shaped automatic CD-changer with six slots for CDs arranged on a circle. If Juan has six CDs, in how many different arrangements can his six CDs be placed on the CD-changer if the blues CD must be put into slot number 1 and the rock CD into one of the slots next to slot number 1?"
Since the blues CD must be placed in slot 1, the number of options for the blues CD = 1.
Since the rock CD must be in one of the 2 slots adjacent to slot 1, the number of options for the rock CD = 2.
The number of ways to arrange the 4 remaining songs in the remaining 4 slots = 4*3*2*1.
Multiplying the numbers above, we get:
Total number of arrangements = 1*2*4*3*2*1 = 48.

Check here for another question about circular arrangements:

https://www.beatthegmat.com/seating-arra ... 85488.html
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