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Combinations q (1 Viewer)
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ExtremelyBoredUser
Bored Uni Student
(10C5)/2 = 126
This is how I think of these type of questions so take this with a grain of salt, there's many ways to interpret combinatorics questions:
The reason for this is that the teams are indistinguishable rather than distinguishable meaning that each groups can be selected in an identical manner and hence you would do the regular 10C5 (5 people in one group) and divide it by 2 to account for situations where the groups are identical but in different positions.
e.g
Scenario 1:
G1: ABCD G2: EFGH
Scenario 2:
G2: EFGH G1: ABCD
^^^ different order but same combination
If this question asked for 2 distinguishable teams, then you would consider the order of the groups, but the question is simply asking for how they can be split into 2 teams so you would disregard the other scenario with identical combinations but different order. The first question (A) is an example of distinguishable property since you would consider the order of groups; this one you're just only caring about the combinations possible.
This is how I think of these type of questions so take this with a grain of salt, there's many ways to interpret combinatorics questions:
The reason for this is that the teams are indistinguishable rather than distinguishable meaning that each groups can be selected in an identical manner and hence you would do the regular 10C5 (5 people in one group) and divide it by 2 to account for situations where the groups are identical but in different positions.
e.g
Scenario 1:
G1: ABCD G2: EFGH
Scenario 2:
G2: EFGH G1: ABCD
^^^ different order but same combination
If this question asked for 2 distinguishable teams, then you would consider the order of the groups, but the question is simply asking for how they can be split into 2 teams so you would disregard the other scenario with identical combinations but different order. The first question (A) is an example of distinguishable property since you would consider the order of groups; this one you're just only caring about the combinations possible.
For part (i):
Let's just say we want the 2 non-diamond cards first with the diamond card last:
First two cards: 39C2 ways of drawing
Third card: 13C1 ways of drawing
Since this is happening simultaneously, multiply them together. However, this is only 1 way of drawing the three cards so we multiply by 3! since there are 3! ways of drawing these 3 cards.
So number of ways = 39C2.13C1.3! = 57798.
Total number of ways = 52x51x50 = 132600.
.'. probability = 57798/132600 = 741/1700
For part (j):
At least two diamonds
2 diamonds and 1 non-diamond - 13C2*39C1 = 3 024
3 diamonds and 0 non-diamond - 13C3*39C0 = 286
Since both are ways that it can happen, add both together
Total number of ways = 52C3 = 22100
Probability = 3328/22100 = 64/425
I endorse the comments of @ExtremelyBoredUser, you need to think carefully when constructing groups whether they are distinguishable or not.
One way students can be tricked is to be asked in how many ways can a group of ten people be split into:
(a) a group of 4 and a group of 6
(b) two groups of 5
Also, on probabilities, sometimes they are helpful as a first step.
For example, in how many ways can I a group of 10 people be formed into a line to walk through a doorway such that one particular person, Y, enters before Z but after X.
One way students can be tricked is to be asked in how many ways can a group of ten people be split into:
(a) a group of 4 and a group of 6
(b) two groups of 5
Also, on probabilities, sometimes they are helpful as a first step.
For example, in how many ways can I a group of 10 people be formed into a line to walk through a doorway such that one particular person, Y, enters before Z but after X.