Problem Solving

An engagement team consists of a project manager, team leader, and four consultants. There are 2 candidates
for the position of project manager, 3 candidates for the position of team leader, and 7 candidates for the 4
consultant slots. If 2 out of 7 consultants refuse to be on the same team, how many different teams are possible?

a.25
b.35
c.150
d.210
e.300

Lets directly move to

C: four slots, seven people = 7 * 6 * 5 * 4. With the slot method, you need to then stop and realize that there are 4! ways to rearrange the four people you have chosen. So you need to divide 840 by 24 to get 35 unique teams that can be created. Finally, since 2 of the consultants can't work together, we need to remove these specific teams from the 35 possibilities.


Ill use the slot method here,order doesnt matter at all since all are going in to the same committee.

2*1*5*4 = 5/3
4*3*2*1


35-(5/3) should be our answer for the consultants,rather it is 35-10=25.

Why is this wrong?

dddanny2006 wrote:Finally, since 2 of the consultants can't work together, we need to remove these specific teams from the 35 possibilities.


Ill use the slot method here,order doesnt matter at all since all are going in to the same committee.

2*1*5*4 = 5/3
4*3*2*1


35-(5/3) should be our answer for the consultants,rather it is 35-10=25.

Why is this wrong?

When we're counting we're always dealing with whole numbers: for example, there are never going to be 3.5 possible groups. So, as soon as you get a fraction, you know there's a problem.

The problem with the using the "slot method" is that you've treated 2 different selections as though it's one big selection. If you want to approach the question this way, you need to break it up into two parts:

1) selecting 2 who don't get along; and
2) selecting 2 others who do.

For the first part, we have 2 people and we're selecting both of them, so there's 1 possibility. Or, using the "slot method":

2*1

which we then divide by 2! to get rid of duplicate arrangements, leaving us with:
2*1/2*1 = 1

For the second part, we're selecting 2 out of 5. Using the "slot method", we get:

5*4

and again we divide by 2! to acknowledge we could flip these two around, giving us:

5*4/2*1 = 10

Since we're selecting a group of 2 AND another group of 2, we multiply to get:

1*10 = 10 "bad" groups.

Accordingly, the final answer is 35-10 = 25

I hope that clears it up!

Stuart

Thanks for that Stuart.What if we just had to pick 4 consultants out of the 7?In that case we can treat it like one big selection.

7*6*5*4 =35 right?
4!

Im learning new things daily Stuart,I thought that I could directly make a big selection via the slot method,but you corrected that,thanks.New things daily,I get frustrated Stuart,different problems have different methods and further changes.How can I master these?


Steve tell me something about the following problem-

A four digit safe code does not contain the digits 1 and 4 at all. What is the probability that it has at least one even digit?

a) ¼
b) ½
c) ¾
d) 15/16
e) 1/16

Here's my method to do it.

Digits allowed-0,2,3,5,6,7,8,9

No of possible arrangements=8*8*8*8=2096 -----Order matters since its a code but I dont understand what if the code contains the same numbers?

If the code consists of the same letters then the order does not matter,likewise if they're different then order matters.But we directly treat it as a Permutation.Please clear the air Stuart.

Thanks again Stuart.

Stuart Kovinsky wrote:

dddanny2006 wrote:Finally, since 2 of the consultants can't work together, we need to remove these specific teams from the 35 possibilities.


Ill use the slot method here,order doesnt matter at all since all are going in to the same committee.

2*1*5*4 = 5/3
4*3*2*1


35-(5/3) should be our answer for the consultants,rather it is 35-10=25.

Why is this wrong?

When we're counting we're always dealing with whole numbers: for example, there are never going to be 3.5 possible groups. So, as soon as you get a fraction, you know there's a problem.

The problem with the using the "slot method" is that you've treated 2 different selections as though it's one big selection. If you want to approach the question this way, you need to break it up into two parts:

1) selecting 2 who don't get along; and
2) selecting 2 others who do.

For the first part, we have 2 people and we're selecting both of them, so there's 1 possibility. Or, using the "slot method":

2*1

which we then divide by 2! to get rid of duplicate arrangements, leaving us with:
2*1/2*1 = 1

For the second part, we're selecting 2 out of 5. Using the "slot method", we get:

5*4

and again we divide by 2! to acknowledge we could flip these two around, giving us:

5*4/2*1 = 10

Since we're selecting a group of 2 AND another group of 2, we multiply to get:

1*10 = 10 "bad" groups.

Accordingly, the final answer is 35-10 = 25

I hope that clears it up!

Stuart

Thanks for that Stuart.What if we just had to pick 4 consultants out of the 7?In that case we can treat it like one big selection.

7*6*5*4 =35 right?
4!

All you're really doing when you apply the "slot method" to these problem is using the combinations formula:

nCk = n!/k!(n-k)!

When order does NOT matter, that's the formula we use.

So, if we're simply choosing 4 out of 7 and we don't care about order, we get:

7C4 = 7!/4!3! = 7*6*5/3*2*1
(since we can cancel out the 4! on the bottom with part of the 7! above)

or, as you've been writing it:

7C4 = 7!/4!3! = 7*6*5*4/4*3*2*1
(since we could also cancel out the 3!, although that leaves us with more work).

As you can see, the combinations formula is just a modified version of the permutations formula which takes those duplicate groups into account and eliminates them.

Im learning new things daily Stuart,I thought that I could directly make a big selection via the slot method,but you corrected that,thanks.New things daily,I get frustrated Stuart,different problems have different methods and further changes.How can I master these?

When you don't care about order, it's generally simpler just to plug into the combinations formula, applying the formula once for each selection your making.

So, back the "figuring out the bad teams" part of the previous question, we're selecting 2/2 and 2/5, giving us:

2C2 * 5C2 = 1 * 5!/2!3! = 1 * 5*4/2 = 1 * 10 = 10
(remember: when you make MULTIPLE selections, you MULTIPLY the individual selections).

It's also a big time saver to learn the common combinations shortcuts:

nC0=1
nC1=n
nCn=1

for all values of n.

(Stuart) tell me something about the following problem-

A four digit safe code does not contain the digits 1 and 4 at all. What is the probability that it has at least one even digit?

a) ¼
b) ½
c) ¾
d) 15/16
e) 1/16

Here's my method to do it.

Digits allowed-0,2,3,5,6,7,8,9

No of possible arrangements=8*8*8*8=2096 -----Order matters since its a code but I don't understand what if the code contains the same numbers?

Order matters, so it's perfectly OK for the code to contain the same numbers - each code is still unique.

For example, if the correct code is 2388 and you plug in 2833, will the door open? No! So 2388 and 2833 are two unique codes, even though they contain the same 4 numbers, and we have to count each one separately.

If the code consists of the same letters then the order does not matter,likewise if they're different then order matters.But we directly treat it as a Permutation.Please clear the air Stuart.

Thanks again Stuart.

I'm not sure what you mean by this last point. If it's a code, then order always, by definition, matters.

alt approach

PM(2c1) x TL(3c1) x 4 consultants out of which those who didnt want to be in same team put out (5c4)
+
PM(2c1) x TL(3c1) x 4 consultants combining one from those two who doesnt want to stay in same team and rest three from remaining 5 (5c3)
+
PM(2c1) x TL(3c1) x 4 consultants combining another one from those two who doesnt want to stay in same team and rest three from remaining 5 (5c3)
= 150

Alternate approach, slightly roundabout though, but just to add to the # of possible ways of looking at the same thing:

Number of ways of selecting a project manager = 2

Number of ways of selecting a team leader = 3

Now, the consultant part:

Let us count one of the two fussy people(say A) out so that we have 6 people to select from.

i.e. 6C4 = 15

Now let us count the other person, say B, out and include A in. Again we have six people to select from.
i.e. 6C4 = 15

now, total number of 4-people group where A & B are not together = 15 x 2 = 30

But here we have to account for redundancy.

Therefore, we will subtract 5C4(Group of five people that we have counted with both A & B) from 30.

i.e. 30-5=25

So the number of ways a team can be selected is= 2 x 3 x 25 = 150.