Please give me a full solution: Chuck has 9 colored beans: 6 blue beans, one green, one red and one yellow. He wants to arrange these beans such that on each side of a non-blue bean there's a blue one. In how many ways can he do that ? Thanks in advance ..

56ways

Step-by-step explanation:

given data:

total beans = 9

blue beans = 6

red beans = 1

yellow beans = 1

green beans = 1

solution:

This is a permutation problem because there is a condition attached to it which states that the beans most be arranged in a way that atleast each none blue has a blue bean on its side when shared.

= ( r + n - 1 ! ) / r ! ( n - 1 )!

= 3 + 6 - 1 / 3! ( 6-1 )!

= 8! / 3! x 5!

= 40320/ 720

= 56 ways

step-by-step explanation:

; fx

replace   with   27   and simplify.

10 , 27 , 10 , 70 , 10 , 35 , 10 , 48

step-by-step explanation: