Question 1207218:
In how many ways can you choose three flags from a collection of seven different flags?
Put your answer in the form [XX]
Found 2 solutions by saradeitz8324, math_tutor2020:
Answer by saradeitz8324(3)   (Show Source):
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
Answer: 35
Explanation
There are 7 choices for the first slot, 6 for the next, and 5 for the final slot.
If order mattered then we'd have 7*6*5 = 210 different permutations.
However, order doesn't matter in this case.
Note there are 3*2*1 = 6 ways to rearrange any trio of flags.
So we have 210/6 = 35 different combinations.
This value is found in Pascal's Triangle. Look at the row that starts with "1,7,..."
We have reached the answer so we can stop here.
The next section is optional.
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Let's label 7 the flags as such
A,B,C,D,E,F,G
Then using this calculator
https://www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html
we can generate this list
| 1 | 2 | 3 | 4 | 5 | | 1 | ABC | ABD | ABE | ABF | ABG | | 2 | ACD | ACE | ACF | ACG | ADE | | 3 | ADF | ADG | AEF | AEG | AFG | | 4 | BCD | BCE | BCF | BCG | BDE | | 5 | BDF | BDG | BEF | BEG | BFG | | 6 | CDE | CDF | CDG | CEF | CEG | | 7 | CFG | DEF | DEG | DFG | EFG |
The table has 7 rows and 5 columns to represent 7*5 = 35 different flag trios, where order doesn't matter.
Order doesn't matter because a group like ABC is the same as BCA.
Once again, listing out these combos is entirely optional.
I don't recommend doing it by hand, and instead suggest using software.
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