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Question Number 179699 by mr W last updated on 01/Nov/22

20 students are numbered with the numbers from 1 to 20. 10 students are randomly selected. what is the probability that the sum of their numbers is exactly 100?

$$\mathrm{20}\:{students}\:{are}\:{numbered}\:{with}\:{the} \\ $$$${numbers}\:{from}\:\mathrm{1}\:{to}\:\mathrm{20}.\:\mathrm{10}\:{students}\:{are} \\ $$$${randomly}\:{selected}.\:{what}\:{is}\:{the} \\ $$$${probability}\:{that}\:{the}\:{sum}\:{of}\:{their} \\ $$$${numbers}\:{is}\:{exactly}\:\mathrm{100}? \\ $$

Commented by Rasheed.Sindhi last updated on 01/Nov/22

■∣■∣■∣■∣■∣■∣■∣■∣■∣■∙∙∙100■′s & 99∣′s Now we′ve to select 9 bars(divisions) (or we′ve to put 9 divisions(bars) ) in a particular way which fulfills your restrictions... I remember this way or some type like this which was introduced by sir prakash jain/mr W sir... perhaps this method helps in such questions...although I′m not sure.

$$\blacksquare\mid\blacksquare\mid\blacksquare\mid\blacksquare\mid\blacksquare\mid\blacksquare\mid\blacksquare\mid\blacksquare\mid\blacksquare\mid\blacksquare\centerdot\centerdot\centerdot\mathrm{100}\blacksquare'\boldsymbol{\mathrm{s}}\:\&\:\mathrm{99}\mid'\boldsymbol{\mathrm{s}} \\ $$$$\mathrm{Now}\:\mathrm{we}'\mathrm{ve}\:\mathrm{to}\:\mathrm{select}\:\mathrm{9}\:\mathrm{bars}\left(\mathrm{divisions}\right) \\ $$$$\left(\mathrm{or}\:\mathrm{we}'\mathrm{ve}\:\mathrm{to}\:\mathrm{put}\:\mathrm{9}\:\mathrm{divisions}\left(\mathrm{bars}\right)\:\right) \\ $$$$\mathrm{in}\:\mathrm{a}\:\mathrm{particular}\:\mathrm{way}\:\mathrm{which}\:\mathrm{fulfills}\:\mathrm{your} \\ $$$$\mathrm{restrictions}... \\ $$$$\mathrm{I}\:\mathrm{remember}\:\mathrm{this}\:\mathrm{way}\:\mathrm{or}\:\mathrm{some}\:\mathrm{type}\:\mathrm{like} \\ $$$$\mathrm{this}\:\mathrm{which}\:\mathrm{was}\:\mathrm{introduced}\:\mathrm{by}\:\mathrm{sir} \\ $$$$\boldsymbol{\mathrm{prakash}}\:\boldsymbol{\mathrm{jain}}/\boldsymbol{{mr}}\:\boldsymbol{{W}}\:{sir}... \\ $$$$\mathrm{perhaps}\:\mathrm{this}\:\mathrm{method}\:\mathrm{helps}\:\mathrm{in}\:\mathrm{such} \\ $$$$\mathrm{questions}...{although}\:{I}'{m}\:{not}\:{sure}. \\ $$$$ \\ $$

Commented by mr W last updated on 01/Nov/22

the difficulty is that the sum of the 10 numbers should be 100. with “stars and bars” method we can not treat the sum of the numbers.

$${the}\:{difficulty}\:{is}\:{that}\:{the}\:{sum}\:{of}\:{the} \\ $$$$\mathrm{10}\:{numbers}\:{should}\:{be}\:\mathrm{100}.\:{with}\:``{stars} \\ $$$${and}\:{bars}''\:{method}\:{we}\:{can}\:{not}\:{treat}\:{the} \\ $$$${sum}\:{of}\:{the}\:{numbers}. \\ $$

Commented by JDamian last updated on 01/Nov/22

p=((combinations suming up 100)/(all combinations)) ?

$${p}=\frac{{combinations}\:{suming}\:{up}\:\mathrm{100}}{{all}\:{combinations}}\:? \\ $$

Commented by mr W last updated on 01/Nov/22

this question is similar to Q179420. but due to the large number 100, we can not enumerate all combinations which give the sum 100.

$${this}\:{question}\:{is}\:{similar}\:{to}\:{Q}\mathrm{179420}. \\ $$$${but}\:{due}\:{to}\:{the}\:{large}\:{number}\:\mathrm{100},\:{we} \\ $$$${can}\:{not}\:{enumerate}\:{all}\:{combinations} \\ $$$${which}\:{give}\:{the}\:{sum}\:\mathrm{100}. \\ $$

Commented by Rasheed.Sindhi last updated on 01/Nov/22

Sir if we put 9 bars between 100 stars then will the sum of ten groups (made by bars) not be 100? Execuse me if I not get to you.

$$\boldsymbol{{Sir}}\:{if}\:{we}\:{put}\:\mathrm{9}\:{bars}\:{between}\:\mathrm{100}\:{stars} \\ $$$${then}\:{will}\:{the}\:{sum}\:{of}\:{ten}\:{groups}\:\left({made}\right. \\ $$$$\left.{by}\:{bars}\right)\:{not}\:{be}\:\mathrm{100}?\: \\ $$$${Execuse}\:{me}\:{if}\:{I}\:{not}\:{get}\:{to}\:{you}. \\ $$

Commented by mr W last updated on 01/Nov/22

this is true. but how can you relate the 100 stars to the numbers from 1 to 20? actually we just want to determine how many solutions following integer equation has: n_1 +n_2 +...+n_(10) =100 with 1≤n_1 <n_2 <...<n_(10) ≤20

$${this}\:{is}\:{true}.\:{but}\:{how}\:{can}\:{you}\:{relate} \\ $$$${the}\:\mathrm{100}\:{stars}\:{to}\:{the}\:{numbers}\:{from}\:\mathrm{1} \\ $$$${to}\:\mathrm{20}? \\ $$$${actually}\:{we}\:{just}\:{want}\:{to}\:{determine} \\ $$$${how}\:{many}\:{solutions}\:{following}\:{integer} \\ $$$${equation}\:{has}: \\ $$$${n}_{\mathrm{1}} +{n}_{\mathrm{2}} +...+{n}_{\mathrm{10}} =\mathrm{100} \\ $$$${with}\:\mathrm{1}\leqslant{n}_{\mathrm{1}} <{n}_{\mathrm{2}} <...<{n}_{\mathrm{10}} \leqslant\mathrm{20} \\ $$

Commented by Rasheed.Sindhi last updated on 01/Nov/22

Yes sir you′re right!

$${Yes}\:{sir}\:{you}'{re}\:{right}! \\ $$

Commented by mr W last updated on 01/Nov/22

using generating function we find out 5095 ways to select 10 numbers whose sum is 100. so the probability is p=((5095)/C_(10) ^(20) )=((5095)/(184756))≈2.76%

$${using}\:{generating}\:{function}\:{we}\:{find}\:{out} \\ $$$$\mathrm{5095}\:{ways}\:{to}\:{select}\:\mathrm{10}\:{numbers}\:{whose} \\ $$$${sum}\:{is}\:\mathrm{100}.\:{so}\:{the}\:{probability}\:{is} \\ $$$${p}=\frac{\mathrm{5095}}{{C}_{\mathrm{10}} ^{\mathrm{20}} }=\frac{\mathrm{5095}}{\mathrm{184756}}\approx\mathrm{2}.\mathrm{76\%} \\ $$

Commented by mr W last updated on 01/Nov/22

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