0
What else can I help you with?
What are Harvard's university living arrangements?
Harvard has several rooming arrangements availableÊfor students. Students can live in gender neutral housing or in married and family housing.
If the teacher plans to assign 8 students to 8 desks for debate. How many different seating arrangements are possible?
Factorial 8, ie 8 x 7 x 6 x 5 x 4 x 3 x 2 = 40320
Ten elementary school students are eligible to be appointed to two positions attendace taker and lunch counter How many unique arrangements of these two positions are possible?
Apex: Permutations
Describe the health and safety arrangements for school exam rooms?
There are no special health and safety arrangements for school rooms in which students take examinations.
A group of 10 students consists of 3 girls and 7 boys how many ways can the students be arranged from left to right if the first 3 students are girls and the remaining students are all boys?
To arrange the 3 girls into the first 3 places:The first place can be any one of 3 girls.The second place can be either of the remaining 2 girls.The number of different possible arrangements of the first 3 places is (3 x 2) = 6In a similar way, we can show that the number of different ways to line up the 7 boys is(7 x 6 x 5 x 4 x 3 x 2) = 5,040 .So the total number of arrangements of all 10 students is (6 x 5,040) = 30,240
How many different committees can be formed from 6 teachers and 33 students if the committee consists of 3 teachers and 2 students?
There are 10560 possible committees.
How many possible ways can a teacher do when she wants assign 4 different tasks to her 4 students?
24
Explain the possible reasons why the actual performance might be different from the budgeted performance.?
TARC AFA students?
Explain the possible reasons why the actual performance might be different from the budgeted performance?
TARC AFA students?
How many ways can a committeee of 6 be chosen from 5 teachers and 4 students if the committee must includes three teachers and three students?
There are 10 different sets of teachers which can be combined with 4 different sets of students, so 40 possible committees.
How many different committees can be formed from 10 teachers and 30 students if the committee consists of 2 teachers and 2 students?
There are (10 x 9)/2 = 45 different possible pairs of 2 teachers. For each of these . . .There are (30 x 29)/2 = 435 different possible pairs of students.The total number of different committees that can be formed is (45 x 435) = 19,575 .
How many different committees can be formed from 11 teachers and 48 students?
To determine the number of different committees that can be formed from 11 teachers and 48 students, we need to clarify the size of the committee and whether there are any restrictions on the selection. If we assume that any combination of teachers and students can be chosen without restrictions, the total number of possible combinations is (2^{11} \cdot 2^{48} = 2^{59}). This accounts for every possible subset of teachers and students, including the empty committee.
