In the design of an electromechanical product, seven differentcomponents are to
ID: 2953322 • Letter: I
Question
In the design of an electromechanical product, seven differentcomponents are to be stacked into a cylindrical casig that holds 12components in a manner that minimizes the impact of shocks. One end of the casing is designated as the bottom and the other endis the top. a.) How many different designs arepossible? b.) If the seven components are allidentical, how many different designs are possible? c.) If the seven components consists of threeof one type of component and four of another type, how manydifferent designs are possible? In the design of an electromechanical product, seven differentcomponents are to be stacked into a cylindrical casig that holds 12components in a manner that minimizes the impact of shocks. One end of the casing is designated as the bottom and the other endis the top. a.) How many different designs arepossible? b.) If the seven components are allidentical, how many different designs are possible? c.) If the seven components consists of threeof one type of component and four of another type, how manydifferent designs are possible?Explanation / Answer
A simple way to think about this problem is that firstfor the sake of clarity think of all the 12components(including 5 empty spaces) as 12 different objects andnow we can arrange this in 12! Ways. This is because firstslot can be filled in 12 ways second one in 11 ways and so on..
But of the 12! Ways 5 objects are identical and hence the 5!Combination are actually only 1 combination hence only 12!/5! Areunique ones. Therefore total number of ways = 3991680.
b) If out of these 12 objects all 7objects are of one type and all 5 empty spaces are of one type bythe logic we have counted a single object 5!*7! More number oftimes . So total # of ways of arranging is 12!/(5!*7!) =792
c) If out of the 7 objects 3are of one kind and 4 are of other then total number of ways =12!(5!*3!*4!) = 27720
One easy way to convince yourself is to take an example .
A1 a2 b1 can be arranged in 3! Ways . now if you consider a1 and a2as identical only 3 ways of arranging. This logic seems to be a bitdifficult to internalize. Also another good way is to first thinkof them as identical and then each combination gives n! ways wheren = number of such objects earlier considered as identical.
Hope this helps . Feel free to ask for any clarification