shelrod007 wrote:
I not able to understand the solution on the Mangoosh blog ..
Over the course of a full-day seminar, 66 students must give short presentations. Most of the presentations are independent, but Sam’s presentation refers to something in Ruth’s, so Ruth must go before Sam; similarly, Matt’s presentation refers to something in Ruth’s and something in Sam’s, so Matt’s must come after both of those. These three presentations need not be consecutive with each other: whatever the order is, Ruth’s must come before the other two but not necessarily first of the 66, and sometime later, Matt must come after the other two but not necessarily last of all. The other 63 participants can present in any order. How many orders obey these constraints?
(A) 22!
(B) 63!
(C) 63!/3!
(D) 66!/3
(E) 66!/3!
Link :
https://magoosh.com/gmat/2015/counting-practice-problems-for-the-gmat/hi shelrod..
lets take a smaller value , 4 A, B, C and D.. there are 4 people in which there is a requirement of A speaking before B and rest can come anywhere..
so it can be ..A..B.. but not ..B..A..
now if we fix two others say at 1st and 2nd.. we will have C,D,A,B and C,D,B,A, out of these only one is correct.. similarily all other scenarios will have one case where A will be befor B and one wher B before A..so only one will be correct..
in this Q, there are three people where restrictions have been put.. R>S>M...rest can be anywhere..
if u take a fixed position for remaining 63, these three people can be placed in 3! ways amongst each other. out of these 3!, only one is correct where r>s>m..
similarily for all possible 66! ways, in which 63 people can be fixed and these 3 people can be fixed only in 1 way out of 3! possible ways..
this is the reason we divide 66! by 3! to cater for the restriction put..
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