MHB Number of Possible Sums with 1-15 Cards

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The discussion centers on calculating the number of distinct sums possible with a set of cards numbered 10, 100, and 500. The initial answer proposed is 143, but some participants argue that at least two cards are needed for a sum, leading to a different calculation method involving combinations. One user claims to have found 141 distinct sums and questions which sums might be missing to reach the total of 143. There is also mention of a formula for calculating distinct sums, but clarity on its application in this context remains unresolved. The conversation highlights confusion over the correct interpretation of the problem and the calculation methods used.
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There are 8 cards with number 10 on them, 5 cards with number 100 on them and 2 cards with number 500 on them. How many distinct sums are possible using from 1 to all of the 15 cards?Answer given is 143. But my logic is for any sum, at least 2 numbers are needed. So, there are $\binom{15} {2} + \binom{15}{3}+...\binom{15}{15} $ distinct sums. So, I think answer 143 is wrong.
What is your opinion?
 
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\begin{pmatrix}15 \\ 2 \end{pmatrix} is the number of sums of two of the numbers, \begin{pmatrix}15 \\ 3 \end{pmatrix} is the number of sums of three of the numbers, etc. But they won't be distinct sums.
 
Dhamnekar Winod said:
There are 8 cards with number 10 on them, 5 cards with number 100 on them and 2 cards with number 500 on them. How many distinct sums are possible using from 1 to all of the 15 cards?Answer given is 143. But my logic is for any sum, at least 2 numbers are needed. So, there are $\binom{15} {2} + \binom{15}{3}+...\binom{15}{15} $ distinct sums. So, I think answer 143 is wrong.
What is your opinion?

The directions clearly state from 1 to 15, so their solution is based on that premise.
 
CountryBoy said:
\begin{pmatrix}15 \\ 2 \end{pmatrix} is the number of sums of two of the numbers, \begin{pmatrix}15 \\ 3 \end{pmatrix} is the number of sums of three of the numbers, etc. But they won't be distinct sums.

Hello,

I have computed 141 distinct sums. But the answer is 143. Which 2 distinct sums i omitted, would you tell me? Is the answer 143 wrong? All the 141 distinct sums are

20,30,40,50,60,70,80,110,120,130,140,150,160,170,180,200,210,220,230,240,250,260,270,280,300,310,320,330,340,350,360,370,380,400,410,420,430,440,450,460,470,480,500,510,520,530,540,550,560,570,580,600,610,620,630,640,650,660,670,680,700,710,720,730,740,750,760,770,780,800,810,820,830,840,850,860,870,880,900,910,920,930,940,950,960,970,980,1000,1010,1020,1030,1040,1050,1060,1070,1080,1100,1110,1120,1130,1140,1150,1160,1170,1180,1200,1210,1220,1230,1240,1250,1260,1270,1280,1300,1310,1320,1330,1340,1350,1360,1370,1380,1400,1410,1420,1430,1440,1450,1460,1470,1480,1500,1510,1520,1530,1540,1550,1560,1570,1580.


I know one formula $\binom{r+n-1}{r-n+1}$ which computes distinct sums, where n=cells and r= balls. How to use that here? Or is there any other formula?
 
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