A committee must be formed with 4 teachers and 4 students. If there are 6 teachers to choose from, and 12 students, how many different ways could the committee be made?
gracias
i dont get it
what is
C(6,4) * C(12,4) ?
do i multiply 15 times 495?
7425
Gracias por tu pregunta tan interesante. Hmm, vamos a ver... Si tienes 6 maestros y necesitas elegir 4 de ellos, eso se puede calcular aplicando una fórmula llamada "combinatoria". La combinatoria de 6 eligiendo 4 es igual a 6!/(4!(6-4)!), lo cual nos da un total de 15 combinaciones posibles para los maestros.
Ahora, para los estudiantes, tienes 12 y necesitas elegir 4 de ellos. De manera similar, la combinatoria de 12 eligiendo 4 es igual a 12!/(4!(12-4)!), lo cual nos da un total de 495 combinaciones posibles para los estudiantes.
Para saber el total de combinaciones posibles de formar el comité con 4 maestros y 4 estudiantes, simplemente multiplicamos el número de combinaciones de maestros por el número de combinaciones de estudiantes. Así que, 15 x 495 = 7425.
Por lo tanto, hay un total de 7425 maneras diferentes de formar el comité. ¡Espero que mi respuesta te haya sacado una sonrisa mientras resolvemos matemáticas! ¿Algún otro acertijo o pregunta? Estoy aquí para divertirte y ayudarte.
To get the number of different ways the committee can be formed, we can use the concept of combinations.
The number of ways to choose 4 teachers from a group of 6 can be calculated using the formula for combinations:
C(6, 4) = 6! / (4! * (6-4)!)
where "!" denotes factorial.
Simplifying this expression, we get:
C(6, 4) = (6 * 5 * 4!) / (4! * 2!)
The factorial terms in the numerator and the denominator cancel out, leaving:
C(6, 4) = (6 * 5) / (2 * 1) = 15
So, there are 15 different ways to choose 4 teachers from a group of 6.
Similarly, the number of ways to choose 4 students from a group of 12 can be calculated using the same formula:
C(12, 4) = 12! / (4! * (12-4)!)
Simplifying this expression, we get:
C(12, 4) = (12 * 11 * 10 * 9 * 8 * 7 * 6!) / (4! * 8 * 6!)
The factorial terms and 6! terms in the numerator and denominator cancel out, leaving:
C(12, 4) = (12 * 11 * 10 * 9) / (4 * 3 * 2 * 1) = 495
So, there are 495 different ways to choose 4 students from a group of 12.
To find the total number of different ways the committee can be formed, we multiply the number of ways to choose the teachers and students together:
Total ways = C(6, 4) * C(12, 4)
Substituting the values we calculated:
Total ways = 15 * 495 = 7425
Therefore, there are 7425 different ways the committee can be formed.
¡De nada! Espero que esto ayude. Si tienes más preguntas, no dudes en hacerlas.
C(6,4) is the notation we use to find the number of ways you can CHOOSE 4 from 6 items
if the order does not matter. This is called a combination.
C(n,r) = n!/(r!9n-r)!)
sometimes written as nCr
If you are getting these kind of questions, you MUST be familiar with this terminology.