A probability problem by Prithwish Roy
Let n 1 < n 2 < n 3 < n 4 < n 5 be positive integers such that n 1 + n 2 + n 3 + n 4 + n 5 = 2 0 . Then find the number of such distinct arrangements ( n 1 , n 2 , n 3 , n 4 , n 5 ) .
The answer is 7.
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2 solutions
Will you please present it as your solution. @Prithwish Roy
Correct but it can be done more easily by bijective transformation
Let m i = n i − i therefore we have to find no of solutions of m 1 + m 2 + m 3 + m 4 + m 5 = 5 where m i can be any non negative integer (including 0) and the elements can be equal too. Now it can be easily seen the solutions are (0,5) (2,3) (4,1) (1,1,3) (1,1,1,1,1) (1,1,1,2)
@Kushal Bose how can this problem be solved by stars and bars?
2+3+4+5+6 = 20.
So 1st term should be ≤ 2
Similarly finding cases we get (1,2,3,4,10),(1,2,3,5,9),(1,2,3,6,8),(1,2,4,6,7),(1,2,4,5,8),(1,3,4,5,7),(2,3,4,5,6). As solutions.