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Question Number 21933 by Tinkutara last updated on 07/Oct/17

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  = 20. Then the number of such  distinct arrangements (n_1 , n_2 , n_3 , n_4 , n_5 )  is

Letn1<n2<n3<n4<n5bepositive integerssuchthatn1+n2+n3+n4+ n5=20.Thenthenumberofsuch distinctarrangements(n1,n2,n3,n4,n5) is

Commented bymrW1 last updated on 08/Oct/17

7 ?  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  2  4  5  6  ⇒ should be 1  3  4  5  7  2  3  4  5  6

7? 123410 12359 12368 12458 12467 12456shouldbe13457 23456

Commented byRasheed.Sindhi last updated on 08/Oct/17

1+3+4+5+7

1+3+4+5+7

Commented byRasheed.Sindhi last updated on 08/Oct/17

line number last but one  1  2  4  5  6  1 + 2 + 4 + 5 + 6 =18?  If this line is replaced by  1  3  4  5  7  the answer is 7 yet

linenumberlastbutone 12456 1+2+4+5+6=18? Ifthislineisreplacedby 13457 theansweris7yet

Commented byTinkutara last updated on 08/Oct/17

mrW1, why you don′t take 1+3+4+5+7?  But your answer 7 is correct.

mrW1,whyyoudonttake1+3+4+5+7? Butyouranswer7iscorrect.

Commented bymrW1 last updated on 08/Oct/17

Thanks to you both! I made a mistake.  1  3  4  5  7  is correct, not 1  2  4  5  6.

Thankstoyouboth!Imadeamistake. 13457iscorrect,not12456.

Commented bymrW1 last updated on 09/Oct/17

Certainly we don′t need to list all the  concrete solutions if we only need to  know how many solutions exist. Besides  for large numbers, for example 100 instead  of 20, it′s not easy to list all the possible  solutions.    We need to know the number of solutions for  n_1 +n_2 +n_3 +n_4 +n_5 =20  under the condition  1≤n_1 <n_2 <n_3 <n_4 <n_5     if we replace  n_1 =a_1 +1  n_2 =a_2 +2  n_3 =a_3 +3  n_4 =a_4 +4  n_5 =a_5 +5  then  the equation n_1 +n_2 +n_3 +n_4 +n_5 =20  will change to a_1 +a_2 +a_3 +a_4 +a_5 =5  and  the condiction 1≤n_1 <n_2 <n_3 <n_4 <n_5   will change to 0≤a_1 ≤a_2 ≤a_3 ≤a_4 ≤a_5     but  the number of solutions of  a_1 +a_2 +a_3 +a_4 +a_5 =5  under the condition  0≤a_1 ≤a_2 ≤a_3 ≤a_4 ≤a_5   means the number of partitions of  the integer 5 which is denoted by  P(5).    P(5)=7 ⇒ number of solutions is 7.    the 7 partitions of the integer 5 are:  a_1 + a_2  + a_3  + a_4  + a_5     ⇒   n_1     n_2    n_3    n_4     n_5   0       0        0        0        5               1       2     3     4     10  0       0        0        1        4               1       2     3     5     9  0       0        0        2        3               1       2     3     6     8  0       0        1        1        3               1       2     4     5     8  0       0        1        2        2               1       2     4     6     7  0       1        1        1        2               1       3     4     5     7  1       1        1        1        1               2       3     4     5     6

Certainlywedontneedtolistallthe concretesolutionsifweonlyneedto knowhowmanysolutionsexist.Besides forlargenumbers,forexample100instead of20,itsnoteasytolistallthepossible solutions. Weneedtoknowthenumberofsolutionsfor n1+n2+n3+n4+n5=20 underthecondition 1n1<n2<n3<n4<n5 ifwereplace n1=a1+1 n2=a2+2 n3=a3+3 n4=a4+4 n5=a5+5 then theequationn1+n2+n3+n4+n5=20 willchangetoa1+a2+a3+a4+a5=5 and thecondiction1n1<n2<n3<n4<n5 willchangeto0a1a2a3a4a5 but thenumberofsolutionsof a1+a2+a3+a4+a5=5 underthecondition 0a1a2a3a4a5 meansthenumberofpartitionsof theinteger5whichisdenotedby P(5). P(5)=7numberofsolutionsis7. the7partitionsoftheinteger5are: a1+a2+a3+a4+a5n1n2n3n4n5 00005123410 0001412359 0002312368 0011312458 0012212467 0111213457 1111123456

Commented byTinkutara last updated on 08/Oct/17

Is there any formula for partitions?  How to list P(85)?

Isthereanyformulaforpartitions? HowtolistP(85)?

Commented bymrW1 last updated on 09/Oct/17

there is no formula to calculate the  integer partitions directly. usually we  use generating functions.

thereisnoformulatocalculatethe integerpartitionsdirectly.usuallywe usegeneratingfunctions.

Commented byTinkutara last updated on 09/Oct/17

Got it.

Gotit.

Commented bymrW1 last updated on 09/Oct/17

mr. ajfour had an example in Q21802  for using generating function to find  number of integer solutions. it is not  really a function but a method.

mr.ajfourhadanexampleinQ21802 forusinggeneratingfunctiontofind numberofintegersolutions.itisnot reallyafunctionbutamethod.

Commented bymrW1 last updated on 09/Oct/17

see also Q21800

seealsoQ21800

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