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16 Feb 2021, 03:00
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C
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Let Pn denotes the number of ways of selecting 3 people out of ‘n’ sitting in a row, if no two of them are consecutive and Qn is the corresponding figure when they are in a circle. If Pn – Qn = 6, then ‘n’ is equal to
(A) 8
(B) 9
(C) 10
(D) 12
(E) 13
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
(A) 8
(B) 9
(C) 10
(D) 12
(E) 13
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
18 Feb 2021, 07:43
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Let the persons selected be X, Y and Z as shown below
|- - - X - - - Y - - - Z - - -|
Let a denote the number of persons sitting to the left of X.
Let b denote the number of persons between X and Y.
Let c denote the number of persons between Y and Z.
Let d denote the number of persons to the right of Z.
Therefore a + b + c + d = n - 3 as X,Y and Z are not included in a,b,c and d.
a and d can be 0 assuming that the 1st person is X and the last person is Z.
b and c cannot be 0 as then X and Y or Y and Z would become consecutive. To avoid this we assume b and c to be another variable + 1
Let us assume that b = p + 1 and c = q + 1, where now p and q can be equal to 0.
Now a + p + 1 + q + 1 + d = n - 3
Therefore a + p + q + d = n - 5
If the number of non negative integer solutions for the equation x1+x2 + ..+ xn = nx1+x2 + ..+ xn = n, then the number of ways the distribution can be done is
\(^{n+r−1}C_{r−1}\). In this case, value of any variable can be zero.
Here r = 4 and n = n - 5
Therefore Pn = \(^{n − 5 + 4−1}C_{4−1} = ^{n−2}C_3\)
Now the same principle applies when seated in a circle, but it is a little bit more complex. let us understand it with numbers.
Suppose n = 5 and the same condition is to be applied.
Total number of ways where no 2 are adjacent = Total ways of choosing 3 people - Total ways where all are adjacent - Total ways where 2 are adjacent and 1 is separate
(a) Total ways of choosing 3 people = 5C3. Therefore for n people = nC3
(b) Total ways where all are adjacent = 5 ways. Therefore for n people it is n ways
(c) Total ways where 2 are adjacent and 1 is separate = Any 2 adjacent people can be selected in 5 ways
Any person non adjacent to these 2 can be chosen in 1 way = 5 - 4 ways (this is constant for any value we take for any number)
So total number of ways = 5 * (5 - 4) ways. For n people it is n * (n - 4) ways
Qn = nC3 - n - n(n - 4)
Given that Pn - Qn = 6
\(^{n−2}C_3 - [^{n}C_3 - n - n(n - 4)] = 6\)
\(\frac{(n - 2)!}{(n - 2 - 3)! * 3!} - [\frac{n!}{(n - 3)! * 3!} - n - n^2 + 4n] = 6\)
\(\frac{(n - 2)!}{(n - 5)! * 6} - [\frac{n!}{(n - 3)! * 6} - n^2 +3n] = 6\)
\(\frac{(n - 2)(n - 3)(n - 4)(n -5)!}{(n - 5)! * 6} - [\frac{n(n -1)(n -2) (n - 3)!}{(n - 3)! * 6}] + n^2 - 3n = 6\)
\(\frac{(n - 2)(n - 3)(n - 4)}{6} - [\frac{n(n -1)(n -2) }{6}] + n^2 - 3n = 6\)
\(\frac{(n - 2)}{6} [(n -3)(n -4) - [n(n - 1)] + n^2 - 3n = 6\)
\(\frac{(n - 2)}{6} [n^2 - 7n + 12 – (n^2 – n)] + n^2 - 3n = 6\)
\(\frac{(n - 2)}{6} [n^2 - 7n + 12 - n^2 + n] + n^2 – 3n = 6\)
\(\frac{(n - 2)}{6} (- 6n + 12) + n^2 – 3n = 6\)
\( (n - 2)(-n + 2) + n^2 – 3n = 6\)
\( –n^2 + 2n + 2n - 4 + n^2 – 3n = 6\)
4n - 3n = 10
n = 10
Option C
Arun Kumar
|- - - X - - - Y - - - Z - - -|
Let a denote the number of persons sitting to the left of X.
Let b denote the number of persons between X and Y.
Let c denote the number of persons between Y and Z.
Let d denote the number of persons to the right of Z.
Therefore a + b + c + d = n - 3 as X,Y and Z are not included in a,b,c and d.
a and d can be 0 assuming that the 1st person is X and the last person is Z.
b and c cannot be 0 as then X and Y or Y and Z would become consecutive. To avoid this we assume b and c to be another variable + 1
Let us assume that b = p + 1 and c = q + 1, where now p and q can be equal to 0.
Now a + p + 1 + q + 1 + d = n - 3
Therefore a + p + q + d = n - 5
If the number of non negative integer solutions for the equation x1+x2 + ..+ xn = nx1+x2 + ..+ xn = n, then the number of ways the distribution can be done is
\(^{n+r−1}C_{r−1}\). In this case, value of any variable can be zero.
Here r = 4 and n = n - 5
Therefore Pn = \(^{n − 5 + 4−1}C_{4−1} = ^{n−2}C_3\)
Now the same principle applies when seated in a circle, but it is a little bit more complex. let us understand it with numbers.
Suppose n = 5 and the same condition is to be applied.
Total number of ways where no 2 are adjacent = Total ways of choosing 3 people - Total ways where all are adjacent - Total ways where 2 are adjacent and 1 is separate
(a) Total ways of choosing 3 people = 5C3. Therefore for n people = nC3
(b) Total ways where all are adjacent = 5 ways. Therefore for n people it is n ways
(c) Total ways where 2 are adjacent and 1 is separate = Any 2 adjacent people can be selected in 5 ways
Any person non adjacent to these 2 can be chosen in 1 way = 5 - 4 ways (this is constant for any value we take for any number)
So total number of ways = 5 * (5 - 4) ways. For n people it is n * (n - 4) ways
Qn = nC3 - n - n(n - 4)
Given that Pn - Qn = 6
\(^{n−2}C_3 - [^{n}C_3 - n - n(n - 4)] = 6\)
\(\frac{(n - 2)!}{(n - 2 - 3)! * 3!} - [\frac{n!}{(n - 3)! * 3!} - n - n^2 + 4n] = 6\)
\(\frac{(n - 2)!}{(n - 5)! * 6} - [\frac{n!}{(n - 3)! * 6} - n^2 +3n] = 6\)
\(\frac{(n - 2)(n - 3)(n - 4)(n -5)!}{(n - 5)! * 6} - [\frac{n(n -1)(n -2) (n - 3)!}{(n - 3)! * 6}] + n^2 - 3n = 6\)
\(\frac{(n - 2)(n - 3)(n - 4)}{6} - [\frac{n(n -1)(n -2) }{6}] + n^2 - 3n = 6\)
\(\frac{(n - 2)}{6} [(n -3)(n -4) - [n(n - 1)] + n^2 - 3n = 6\)
\(\frac{(n - 2)}{6} [n^2 - 7n + 12 – (n^2 – n)] + n^2 - 3n = 6\)
\(\frac{(n - 2)}{6} [n^2 - 7n + 12 - n^2 + n] + n^2 – 3n = 6\)
\(\frac{(n - 2)}{6} (- 6n + 12) + n^2 – 3n = 6\)
\( (n - 2)(-n + 2) + n^2 – 3n = 6\)
\( –n^2 + 2n + 2n - 4 + n^2 – 3n = 6\)
4n - 3n = 10
n = 10
Option C
Arun Kumar
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07 Jan 2025, 13:27
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Bunuel
This was a tough one with lengthy calculations -
Pn = Total number of ways of selecting 3 people out of n - No. of ways when 3 of them are together - No. of ways when 2 of them are together(Have to take 2 cases - when these 2 people are in the middle + when these 2 people are at the front or bottom)
\(nC3 - (n-2) - ((n-3)*(n-4)C1 + 2*(n-3)C1)\)
\(nC3 - (n-2) - (n-3)*(n-2)\)
\(nC3 - (n-2)*(n-2)\)
\(\frac{n*(n-1)*(n-2)}{6} - (n-2)^2\)
\(\frac{(n-2)}{6}*(n^2 - 7n + 12)\)
\(\frac{(n-2)*(n-3)*(n-4)}{6}\)
Similarly,
Qn = Total number of ways of selecting 3 people out of n - No. of ways when 3 of them are together - No. of ways when 2 of them are together
\(nC3 - n - (n*(n-4)C1)\)
\(nC3 - n - n*(n-4)\)
\(nC3 - n*(n-3)\)
\(\frac{n*(n-1)*(n-2)}{6} - n*(n-3)\)
\(\frac{n}{6}*(n^2 - 9n + 20)\)
\(\frac{n*(n-4)*(n-5)}{6}\)
Pn - Qn = 6
\(\frac{(n-2)*(n-3)*(n-4)}{6} - \frac{n*(n-4)*(n-5)}{6} = 6 \)
\(\frac{n-4}{6}*(n^2 - 5n + 6 - n^2 + 5n) = 6\)
\(\frac{n-4}{6} = 1\)
\(n-4 = 6\)
\(n = 10\)
IMO: C
