Welcome to our community

Be a part of something great, join today!

Rule of sum and rule pf product: How many distinct sums are possible using from 1 to all of the 15 cards

Dhamnekar Winod

Active member
Nov 17, 2018
100
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?
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
368
[tex]\begin{pmatrix}15 \\ 2 \end{pmatrix}[/tex] is the number of sums of two of the numbers, [tex]\begin{pmatrix}15 \\ 3 \end{pmatrix}[/tex] is the number of sums of three of the numbers, etc. But they won't be distinct sums.
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
639
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.
 

Dhamnekar Winod

Active member
Nov 17, 2018
100
[tex]\begin{pmatrix}15 \\ 2 \end{pmatrix}[/tex] is the number of sums of two of the numbers, [tex]\begin{pmatrix}15 \\ 3 \end{pmatrix}[/tex] 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?
 
Last edited: