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.
To find the number of combinations of 5 students that can be chosen from 24 students, you can use the combination formula ( C(n, k) = \frac{n!}{k!(n-k)!} ). In this case, ( n = 24 ) and ( k = 5 ), so the calculation is ( C(24, 5) = \frac{24!}{5!(24-5)!} = \frac{24!}{5! \cdot 19!} = \frac{24 \times 23 \times 22 \times 21 \times 20}{5 \times 4 \times 3 \times 2 \times 1} = 42504 ). Therefore, the teacher can choose from 42,504 different combinations of 5 students.
two
There are 234,345,679 combinations of drinks in he world.
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
There could be many different symbols. It probably looks like a person swimming. It could be different though.
I think there are 88 different combinations of coins that can make up 66 cents.
2 adults, 6 students 4 adults, 3 students 6 adults 9 students 4 combinations
There are 84 different combinations possible for the committee of 6, taken from4 students and 5 teachers.1.- The committee with 4 students has 4C4 number of combinations of 4 students out of 4 and 5C2 number of combinations of 2 teachers out of 5 to be combined with. The product of these two give the number of different combinations possible in a committee formed by 4 students and 2 teachers.2.- The committee with 3 students has 4C3 number of combinations of 3 students out of 4 and 5C3 number of combinations of 3 teachers out of 5 to be combined with. The product of these two give the number of different combinations possible in a committee formed by 3 students and 6 teachers.3.- The committee with 2 students has 4C2 number of combinations of 2 studentsout of 4 and 5C4 number of combinations of 4 teachers out of 5 to be combined with. The product of these two give the number of different combinations possible in a committee formed by 2 students and 4 teachers.4.- The committee with 1 student has 4C1 number of combinations of 1 student out of 4 students and 5C5 number of combinations of 5 teachers out of 5 to be combined with. The product of these two give the number of different combinations possible in a committee formed by 1 student and 5 teachers.We now add up all possible combinations:4C4∙5C2 + 4C3∙5C3 + 4C2∙5C4 + 4C1∙5C5 = 1(10) + 4(10) +6(5) + 4(1) = 84There are 84 different combinations possible for the committee of 6, taken from4 students and 5 teachers.[ nCr = n!/(r!(n-r)!) ][ n! = n(n-1)(n-2)∙∙∙(3)(2)(1) ]
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.