Problem Solving

Hi All,

Can someone please tell what is the difference between the following two questions and how to arrive at the answers of each ?

Q1) Six alphabets P,Q,R,S,T and U have to be arranged in six numbered positions(1-6). How many ways can you arrange them so that P is not in position numbered 3, Q is not in position numbered 5 and R is not in position numbered 6.
a) 714
b) 426
c) 240
d) 120
e) 252
[spoiler]OA = A.[/spoiler]

Q2) Six alphabets P,Q,R,S,T and U have to be arranged in six numbered positions(1-6). How many ways can you arrange them so that neither P is in position numbered 3, nor Q is in position numbered 5 and nor R is in position numbered 6.
a) 714
b) 426
c) 240
d) 120
e) 252
[spoiler]OA = B.[/spoiler]

Q1.

lets fix P, Q and R for 3rd, 5th and 6th positions respectively.the rest of the 3 places can be filled by 3*2*1=6 ways.

total possible arrangements: 6!=720 ways.
of which the above 6 ways are forbidden, so allowed number of arrangments: 720-6=714 ways.

scoobydooby wrote:Q1.

lets fix P, Q and R for 3rd, 5th and 6th positions respectively.the rest of the 3 places can be filled by 3*2*1=6 ways.

total possible arrangements: 6!=720 ways.
of which the above 6 ways are forbidden, so allowed number of arrangments: 720-6=714 ways.

Thanks Scooby.

Can you please help me understand what exactly is the difference that Q1 and Q2 are asking...I am really stumped by their language...looks more or less same to me.

Q1 is putting conditions on position 3, 5, and 6 simultaneously. So for example the following is allowed:

_ _ P _ Q R --> whenever these 3 are in the following positions we need to exclude but you could have the following

_ _ P _ Q S or
_ _ P _ R Q or
_ _ S _ Q P

in Q2, the other 3 conditions are independent. So example 1, 2, 3 above would each not be counted (hence reducing the combinations possible) - as long as one condition is true then we don't count.

One way to think of Q1 is :
(restriction 1)AND(restriction 2)AND(restriction 3)
while Q2 is
(restriction 1)OR(restriction 2)OR(restriction 3)

Hope that helps, though I'm not convinced by answer #2.