Three scenarios: 1 Senior, 2 seniors, and 3 seniors
(1) 1 Senior and 2 Junior
(4 * 6 * 5)/2! = 60
*divide by 2! because of the redundancy of the the two spots for the junior partner*
(2) 2 Senior and 1 Junior
(4 * 3 * 6)/2! = 36
*divide by 2! because of the redundancy of the the two spots for the senior partners*
(3) 3 Seniors and 0 Juniors
(4 * 3 * 2)/3! = 4
*divide by 3! because of the redundancy of the the three spots for the senior partners*
Add (1), (2), and (3) giving
[spoiler]B) 100[/spoiler]
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Gmat Prep Probablity??
Source: Beat The GMAT — Problem Solving |
I understand the three scenerios but I'm not quite sure how you did the permutations. For example you the first was (4*6*5)/ 2! Why did you do that. Explain please for the slow people.
Thanks and I appreciate it.
Thanks and I appreciate it.
Suppose the senior partners are named A, B, C, and D and the junior partners are U, V, W, X, Y, and Z. For scenario 1 (one senior and two juniors) the group {A, U, X}, for example, is equivalent to the group {A, X, U}... each such grouping will have an additional equivalent grouping, so you must divide by 2 (or, equivalently by 2!) to account for the double counting.
In scenario 3 (three seniors), it's a bit more complicated because there are 3! or 6 ways to order each unique group of 3 partners. So you divide by 3!.
What's critical here is that you're dealing with combinations (order of selection doesn't matter), not permutations (where order does matter). The formula for combinations is nCr = n!/[r!(n-r)!]. Consider it this way:
Scenario 1 (1 senior partner and 2 junior partners)
Senior partner: 4C1 = 4!/(1!*3!) = 4
Junior partners: 6C2 = 6!/(2!*4!) = 15
4*15=60
Scenario 2 (2 senior partners and 1 junior partner)
Senior partner: 4C2 = 4!/(2!*2!) = 6
Junior partners: 6C1 = 6!/(1!*5!) = 6
6*6=36
Scenario 3 (3 senior partners)
4C3 = 4!/(3!*1!) = 4
Add 'em up to get 100.
Rey
In scenario 3 (three seniors), it's a bit more complicated because there are 3! or 6 ways to order each unique group of 3 partners. So you divide by 3!.
What's critical here is that you're dealing with combinations (order of selection doesn't matter), not permutations (where order does matter). The formula for combinations is nCr = n!/[r!(n-r)!]. Consider it this way:
Scenario 1 (1 senior partner and 2 junior partners)
Senior partner: 4C1 = 4!/(1!*3!) = 4
Junior partners: 6C2 = 6!/(2!*4!) = 15
4*15=60
Scenario 2 (2 senior partners and 1 junior partner)
Senior partner: 4C2 = 4!/(2!*2!) = 6
Junior partners: 6C1 = 6!/(1!*5!) = 6
6*6=36
Scenario 3 (3 senior partners)
4C3 = 4!/(3!*1!) = 4
Add 'em up to get 100.
Rey
Rey Fernandez
Instructor
Manhattan GMAT
Instructor
Manhattan GMAT
Here is an alternative solution:
First, figure out the [b]total [/b]combinations you can make between Senior and Junior partners: 10! / 3!*(10-3)! = 10! / 3!*7! = 10*9*8 / 3*2 = [b]120[/b]
Then instead of calculating all the permutations where there is at least 1 Senior partner, just calculate the opposite: How many combinations are there with [b]NO [/b]senior partners. This is much simpler since you only need to do a single calculation = 6! / 3!*(6-3)! = 6! / 3!*3! = 6*5*4 / 3*2 = 20
Now that you have those 2 values just substract 120 - 20 = 100.
The rationale is as follows: If you have the total number of combinations and subtract the combinations where there is no seniopr partners, you are left with only with the combinations that have senior partners in them.
Hope that helps!
First, figure out the [b]total [/b]combinations you can make between Senior and Junior partners: 10! / 3!*(10-3)! = 10! / 3!*7! = 10*9*8 / 3*2 = [b]120[/b]
Then instead of calculating all the permutations where there is at least 1 Senior partner, just calculate the opposite: How many combinations are there with [b]NO [/b]senior partners. This is much simpler since you only need to do a single calculation = 6! / 3!*(6-3)! = 6! / 3!*3! = 6*5*4 / 3*2 = 20
Now that you have those 2 values just substract 120 - 20 = 100.
The rationale is as follows: If you have the total number of combinations and subtract the combinations where there is no seniopr partners, you are left with only with the combinations that have senior partners in them.
Hope that helps!
