The first n natural numbers, 1 to n, have to be arranged in a row from left to right. The n numbers are arranged such that there are an odd number of numbers between any two even numbers as well as between any two odd numbers. If the number of ways in which this can be done is 72, then find the value of n?

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Bhargava Vadlamani

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n=17. This is quite tricky. 1 to n natural numbers can be even number of numbers or odd number of numbers. ex - 1 2 3 4 5 or 1 2 3 4 5 6 so here n could be 5 or 6. We don't know hence we consider both the cases. n ---> even . Then there will be n/2 even and n/2 odd numbers in list of n numbers. Now...
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n=17. This is quite tricky. 1 to n natural numbers can be even number of numbers or odd number of numbers. ex - 1 2 3 4 5 or 1 2 3 4 5 6 so here n could be 5 or 6. We don't know hence we consider both the cases. n ---> even . Then there will be n/2 even and n/2 odd numbers in list of n numbers. Now we need observe that when we try to arrange odd number of numbers in between two even or odd numbers we end up having consecutive even and odd number always. So, number of ways will be n/2! * n/2! .Now lets try to equate it to 72. We get nothing as n/2! * n/2! is a perfect square and 72 is not. n -- > odd. Then we will have (n+1)/2 odd numbers and (n-1)/2 even numbers in the list of n numbers. we apply the same logic here too:- (n-1)/2 ! * (n+1)/2 ! = 72 (n^2 - 1)/4 = 72 ---- > n = 17. Hence there are 17 numbers 1 2 3 .... 17. Let me know if you need more explanation. read less
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Correction:- It is (n-1)/2 ! * (N+1)/2 ! So if n is 7 then answer is 144 If n is 5 then answer is 12.
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It needs to be visualized first that there can be only 1 odd number between any two even numbers and vice versa because if you place 3 odd numbers between 2 even numbers then some pair of even numbers will not have any odd numbers between them lets assume a set of numbers 1, 2, 3, 4,...
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It needs to be visualized first that there can be only 1 odd number between any two even numbers and vice versa because if you place 3 odd numbers between 2 even numbers then some pair of even numbers will not have any odd numbers between them lets assume a set of numbers 1, 2, 3, 4, 5, 6 2, 1 ,3 ,5 ,4 ,6 not satisfying the condition thus we have only 2 of types possibility that satisfies the given condition: 1 2 3 4 5 6 2 1 4 3 6 5 the two types of possibility happened because the number of odd and number of even numbers are same but for a set of odd numbers there will be only one type of possibility i.e. starting with odd numbers. e.g.:- 2 3 4 5 1 not satisfying the condition 1 2 3 4 5 Now coming to the question. Lets suppose, n is even and thus n=2t+2 (where t is an integar with t>0) thus t+1 even numbers and t+1 odd numbers 2*(t+1)! (t+1)! = 72 (t+1)! = 6 t=2 n=6. Lets suppose n is odd and thus n=2t+1 (where t is an integar with t>0) thus t even numbers and t+1 odd numbers (2t+1)! (t)! = 72 for t=1 rhs =6 for t=2 rhs =240 thus the above equation will not be satisfied for any value of t. Hence, n = 6 read less
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