There are N people sitting around a circular table. Let’s call them P1, P2, . . . , PN , in order. P1 is sitting between P2 and PN .P2 is sitting between P1 and P3. P3 is sitting between P2 and P4, and so on.
You need to select a non-empty subset of these people, such that no two adjacent people are selected. Find the total number of ways in which you can do so.
For example, suppose N = 3, then we have P1, P2 and P3. If we select P1, then neither P2 nor P3 can be selected. So, the only valid selections are {P1}, {P2}, {P3}.
So there are three possible ways, and hence the answer would be 3.
Suppose N = 4, then we have P1, P2, P3 and P4. The possible subsets are {P1}, {P2}, {P3}, {P4}, {P1, P3}, {P2, P4}. So the answer would be 6.
Find the answer for the following value of N.
Find for : N = 15
Answers
TOTAL NO. OF POSSIBLE SETS IS 198
First, your count for subsets of 22 is off. When you choose
P4, you have already counted pairing that up with
P1 and P2 , and so you only have 66 more options, rather than 77. So, you get 8+8+7+6+5+4+3+2+1=448+8+7+6+5+4+3+2+1=44 options.
[ TOTAL OPTIONS = 44 ]
• Now , for each person we can pick 88 others, giving 11×8=88 options
• Now you need to rule out double-counting: first picking P1 and then picking P6 gives you the same subset as first picking P6 and then picking P1. So, you need to divide by 2
[ 88 \ 2 = 44 ] _______ possible subset of size 2
You can again start with P1, pick P3as a second person, and then have a choice of 6 (P5through P10) as a third. So, that's 6 options. Then, still staying with P1, you can pick P4 as your second person, but now you have only 5 persons left (P6 through P10) for the third one. So, that's another 5 options. Choosing P5 as the seonc gives 4 options ... and continuing this, you get 6+5+4+3+2+1=21 options with P1P1 as the first person. This, of course, repeats for all 11 people, so you get 11⋅21=231 options. But again, you are overcounting, because for any subset of 3, you could have picked any of those 4 as your 'first' person. As such, every triple got counted 3 times, and so you have
[231 \ 3=77 possible subsets of size 3]
Now for subset 4 Again, we start with P1, add P3, then P5, and now you have 4 options left (P7 though P10). Now we can make the third person P6, and that gives you 3+2+1=6
options. Likewise, you get 2+1=3 and 1
one more while having P1 as the second person. So, there are 10+6+3+1=20 subsets with P1 as the first person. Again, multiply by 11 and divide by 4, since each subset gets counted 4 times, since we could have started going around with each of the 4 people in each subset. So: 11× 20/4=55
Now for subset 5 With P1, P3, P5, and P7 you have 2 options left. You can also make P8 the fourth person, and now there is 1 option left. You can make the third P6, and that also leads to 1 options, and finally you can pick P4 as the second, for a final option. Total: 5 subsets with P1. Again, multiply by 11, and divide by size of subsets: 11×5/5=11 possibilities.
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Answer:
TOTAL NO. OF POSSIBLE SETS IS 198
\bold{\large{\underline{\underline{\sf{StEp\:by\:stEp\:explanation:}}}}}
StEpbystEpexplanation:
First, your count for subsets of 22 is off. When you choose
P4, you have already counted pairing that up with
P1 and P2 , and so you only have 66 more options, rather than 77. So, you get 8+8+7+6+5+4+3+2+1=448+8+7+6+5+4+3+2+1=44 options.
[ TOTAL OPTIONS = 44 ]
• Now , for each person we can pick 88 others, giving 11×8=88 options
• Now you need to rule out double-counting: first picking P1 and then picking P6 gives you the same subset as first picking P6 and then picking P1. So, you need to divide by 2
[ 88 \ 2 = 44 ] _______ possible subset of size 2
\bold \green{for \: \:subset \: 3 \implies}forsubset3⟹
You can again start with P1, pick P3as a second person, and then have a choice of 6 (P5through P10) as a third. So, that's 6 options. Then, still staying with P1, you can pick P4 as your second person, but now you have only 5 persons left (P6 through P10) for the third one. So, that's another 5 options. Choosing P5 as the seonc gives 4 options ... and continuing this, you get 6+5+4+3+2+1=21 options with P1P1 as the first person. This, of course, repeats for all 11 people, so you get 11⋅21=231 options. But again, you are overcounting, because for any subset of 3, you could have picked any of those 4 as your 'first' person. As such, every triple got counted 3 times, and so you have
[231 \ 3=77 possible subsets of size 3]
\bold \green{for \: \:subset \: 4 \implies}forsubset4⟹
Now for subset 4 Again, we start with P1, add P3, then P5, and now you have 4 options left (P7 though P10). Now we can make the third person P6, and that gives you 3+2+1=6
options. Likewise, you get 2+1=3 and 1
one more while having P1 as the second person. So, there are 10+6+3+1=20 subsets with P1 as the first person. Again, multiply by 11 and divide by 4, since each subset gets counted 4 times, since we could have started going around with each of the 4 people in each subset. So: 11× 20/4=55
\bold \green{for \: \:subset \: 5 \implies}forsubset5⟹
Now for subset 5 With P1, P3, P5, and P7 you have 2 options left. You can also make P8 the fourth person, and now there is 1 option left. You can make the third P6, and that also leads to 1 options, and finally you can pick P4 as the second, for a final option. Total: 5 subsets with P1. Again, multiply by 11, and divide by size of subsets: 11×5/5=11 possibilities.
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\sf \blue{total \: no. \: of \: subsets \implies}totalno.ofsubsets⟹
\tt \bold{11+44+77+55+11=198 \: subsets}11+44+77+55+11=198subsets