Jane has six different hamsters for which she has two cages, one red and one blue. She wants to put three hamsters in the red cage and three in the blue cage, but two of the hamsters, Felix and Oscar, do not get along and cannot be in the same cage. In how many different ways can she choose which three hamsters to put in the red cage?

12 ways

Step-by-step explanation:

The different ways to choose which three hamsters to put in the red cage.

At first , she must choose one of the two of the hamsters, Felix and Oscar, because they do not get along and cannot be in the same cage.

So, the number of ways to choose one of them = 2

Then she will choose 2 from the remaining 4 hamsters (order does not matter)

The number of ways to choose 2 from 4 = 4C2 = 6

So, the total number of ways = 2 * 6 = 12

75%

step-by-step explanation:

answer: - (a)

(b) 88 seats

step-by-step explanation:

given: in an auditorium, there are 22 seats in the first row and 28 seats in the second row.

thus first term=

second term=

the number of seats in a row continues to increase by 6 with each additional row.

⇒d= 6

a) given formula :

b)   let the number of seats (n) = 12

substitute n=12 in the explicit rule formed in (a)

∴there are 88 seats in 12 rows.