Any combination of 5 students leaves one student out. Since there are 5 possible students to leave out, the number of combinations of all but one student is 5.
The answer is 86 because 100 studets were asked and 14 more students chose soccer than swimming so 100-14=86, then halve 86 to make 43 and 43, then add 43 and 14 to make 57 for soccer. So 57 people chose soccer, and 43 people chose swimming. All together it equals to 100 students.
The number of different groups of students that can be formed from 16 students depends on the size of the groups being formed. If you are looking for all possible combinations of groups of any size (from 1 to 16), you can use the formula for combinations. The total number of combinations would be (2^{16} - 1) (subtracting 1 to exclude the empty group), which equals 65,535 different groups. If you specify a particular group size, the calculation would be different.
two
There are 234,345,679 combinations of drinks in he world.
One bar code can have anywhere from 30 to 70 different combinations within itself. How many combinations are in the world is not known.
The answer is 4,960.
There are 4845 ways to choose 4 people out of 20 20 choose 4 = 20! / (4!16!)
9 combinations - the key person and one of the remaining nine.
The answer is 30C4 = 30*29*28*27/(4*3*2*1) = 27,405
3
The answer is 86 because 100 studets were asked and 14 more students chose soccer than swimming so 100-14=86, then halve 86 to make 43 and 43, then add 43 and 14 to make 57 for soccer. So 57 people chose soccer, and 43 people chose swimming. All together it equals to 100 students.
20
I think there are 88 different combinations of coins that can make up 66 cents.
There could be many different symbols. It probably looks like a person swimming. It could be different though.
2 adults, 6 students 4 adults, 3 students 6 adults 9 students 4 combinations
To calculate the number of ways a committee of 6 can be chosen from 5 teachers and 4 students, we use the combination formula. The total number of ways is given by 9 choose 6 (9C6), which is calculated as 9! / (6! * 3!) = 84. Therefore, there are 84 ways to form a committee of 6 from 5 teachers and 4 students if all are equally eligible.
There can be 4 different non-repeating allele combinations in the gametes of a person with genotype AABBCc: ABC, ACB, BAC, and BCA.