0
What else can I help you with?
In how many different ways can 5 students be arranged for a group picture if all students must face forward and be in a straight line?
For the first spot, you can choose any one of 5 students. For the second spot, you can choose any one of the remaining 4 students. For the third spot, you can choose any one of the remaining 3 students. etc. So the answer is: 5x4x3x2x1 = 120
How many different ways can a teacher select 5 students from a class of 19 to each perform different task?
19*18*17*16*15 = 1,395,360
How many different ways can 10 colored pencils be distributed to a group of 4 art students?
I Dont Know [210]
An algebra class of 21 students must send 5 students to meet with the principal How many different groups of 5 students could be formed from this class?
To find the number of different groups (a) within a larger group (b), use the formula b!/a!.(b-a)! where the ! sign indicates "factorial". In your problem b = 21 and a = 5 so we have 21!/(5!.16!) this simplifies to 21.20.19.18.17/5.4.3.2 cancelling leaves 21.19.3.17 ie 20349
How many ways can a group of 3 students be chosen from a class of 30 students?
Well, honey, there are 30 students in the class, and you want to choose a group of 3. So, you're looking at a classic combination situation. The formula for combinations is nCr = n! / r!(n-r)!, so in this case, it's 30C3 = 30! / 3!(30-3)! = 4060 ways to choose those 3 lucky students. It's like picking the winning lottery numbers, but with fewer tears and more math.
How many different committees can be formed from 9 teachers and 49 students if the committee consists of 2 teachers and 4 students?
Possibilities are (9 x 8)/2 times (49 x 48 x 47 x 46)/24 = 366,121,728/48 =7,627,536 different 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 6 teachers and 49 students if the committee consists of 4 teachers and 4 students?
To determine the number of different committees that can be formed with 4 teachers from 6 and 4 students from 49, we use combinations. The number of ways to choose 4 teachers from 6 is given by ( \binom{6}{4} ), and the number of ways to choose 4 students from 49 is ( \binom{49}{4} ). Thus, the total number of different committees is ( \binom{6}{4} \times \binom{49}{4} ). Calculating this gives ( 15 \times 194580 = 2918700 ) different committees.
A school committee consists of 2 teachers and 4 students. The number of different committees that can be formed from 5 teachers and 10 students is?
To form a committee of 2 teachers from 5, we use the combination formula ( \binom{n}{r} ), where ( n ) is the total number and ( r ) is the number chosen. The number of ways to choose 2 teachers from 5 is ( \binom{5}{2} = 10 ). For the 4 students from 10, the number of ways is ( \binom{10}{4} = 210 ). Therefore, the total number of different committees is ( 10 \times 210 = 2100 ).
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.
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 9 teachers and 41 students if the committee consist of 4 teachers and 2 student?
You can select 4 of the 9 teachers in any order, and for each of those selections you can select 2 of the 41 students in any order. This is two combinations → number_of_ways = ₉C₄ + ₄₁C₂ = 9!/((9-4)!4!) + 41!((41-2)!2!) = 126 + 820 = 946 different committees.
Is this a permutation In how many different ways could a committee of 5 students be chosen from a class of 25 students?
6,375,600
How many different ways can 1 committee of 5 students be selected from a class of 25?
53,130 ways.
How many different committees can be formed from 12 teachers and 36 students if the committee consists of 4 teachers and 3 students?
I think the answer might surprise you! Any 4 from 12 is (12 x 11 x 10 x 9)/(4 x 3 x 2) ie 495; Any 3 from 36 is (36 x 35 x 34)/(3 x 2) ie 7140 These must be multiplied as each teacher set can be combined with each student set giving a total of (deep breath) 3,534,300.
How many 3 member committed can you choose from a group of 18 students?
To determine how many 3-member committees can be formed from a group of 18 students, you can use the combination formula: (C(n, r) = \frac{n!}{r!(n-r)!}), where (n) is the total number of students and (r) is the number of members in the committee. In this case, (n = 18) and (r = 3). Thus, the calculation is (C(18, 3) = \frac{18!}{3!(18-3)!} = \frac{18 \times 17 \times 16}{3 \times 2 \times 1} = 816). Therefore, you can form 816 different 3-member committees from the group of 18 students.
How many ways can a committee of 6 be chosen from 5 teachers and 4 students if all are equally eligible?
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.
