Another Combinatorics Problem

Let n 1 < n 2 < n 3 < n 4 < n 5 n_1 <n_2<n_3<n_4 < n_5 be positive integers such that their sum is 20. Find the number of distinct ordered solutions for ( n 1 , n 2 , n 3 , n 4 , n 5 ) (n_1,n_2,n_3,n_4,n_5) .


The answer is 7.

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4 solutions

Jaiveer Shekhawat
Sep 17, 2015

Puneet Sharma
Oct 6, 2016

pfa!

There are only 7 solutions satisfying n 1 < n 2 < n 3 < n 4 < n 5 n_1<n_2<n_3<n_4<n_5

( 1 , 2 , 3 , 4 , 10 ) , ( 1 , 2 , 3 , 5 , 9 ) , ( 1 , 2 , 3 , 6 , 8 ) , ( 1 , 2 , 4 , 5 , 8 ) , ( 1 , 2 , 4 , 6 , 7 ) , ( 1 , 3 , 4 , 5 , 7 ) , ( 2 , 3 , 4 , 5 , 6 ) (1,2,3,4,10), (1,2,3,5,9), (1,2,3,6,8), (1,2,4,5,8), (1,2,4,6,7), (1,3,4,5,7), (2,3,4,5,6)

Arpan Sarangi
Sep 15, 2015

can you list the actual ordered pairs, please? I only found 2.

michael bye - 5 years, 9 months ago

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Possible solutions are (1, 2, 3, 4, 10) (1, 2, 3, 5, 9) (1, 2, 3, 6, 8) (1, 2, 4, 5, 8) (1, 2, 4, 6, 7) (1, 3, 4, 5, 7) (2, 3, 4, 5, 6) Hence 7 solutions are there.

Arpan Sarangi - 5 years, 8 months ago

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thank you very much. I do these problems at like 3am to 6am and my brain must have shut down thaty night lol

michael bye - 5 years, 8 months ago

Can you please elaborate how come u directly jumped to the conclusion that the number of solutions is 7?

jaiveer shekhawat - 5 years, 9 months ago

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