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email2vm
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permutation and combination questions 03 Aug 2013, 03:47
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I just want to check if there is any better way to solve the below problems:


q1> In how many ways can a person choose 1 or more of 4 apples? Ans: 15

Way1> c(4,1)+c(4,2)+c(4,3)+c(4,4)=15

q2> Three dice are rolled. Find the number of possible outcomes in which at least one dice shows 6?

Way 1> The below ways is time consuming and there should be better way to do this...please sugges

combination 1 of only one 6 : getting 6 in dice 1 and remaining dice may be any number between (1 and 5)
therefore (1X 5X 5)= 25
but it is possible that dice 2 can get 6 or dice 3 can get 6
so 25* 3 =75

Combination 2 of two 6: 1* 1*5= 5 but three such cases again so 5*3= 15

combination 3 of 3 six: 1*1*1 which can be only one case.

Total : 75+15+1= 91

though answer is correct but it is very long procedure, is there any better procedure



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permutation and combination questions 03 Aug 2013, 04:40
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I just want to check if there is any better way to solve the below problems:


q1> In how many ways can a person choose 1 or more of 4 apples? Ans: 15

Way1> c(4,1)+c(4,2)+c(4,3)+c(4,4)=15

q2> Three dice are rolled. Find the number of possible outcomes in which at least one dice shows 6?

Way 1> The below ways is time consuming and there should be better way to do this...please sugges

combination 1 of only one 6 : getting 6 in dice 1 and remaining dice may be any number between (1 and 5)
therefore (1X 5X 5)= 25
but it is possible that dice 2 can get 6 or dice 3 can get 6
so 25* 3 =75

Combination 2 of two 6: 1* 1*5= 5 but three such cases again so 5*3= 15

combination 3 of 3 six: 1*1*1 which can be only one case.

Total : 75+15+1= 91

though answer is correct but it is very long procedure, is there any better procedure
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permutation and combination questions 04 Aug 2013, 00:57
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Q1.) I dont think there is an easier solution. Because the question simply asks the for multiple events. "1 or more". So in essence, they are multiple questions.

Q2.) There is an easier solution. Q'n asks for "at least" one six. therefore you have 3 types of favorable cases right? 1 six or 2 six or 3 six.
BUT remember, Total outcomes = Fav. outcomes + Unfav. outcomes. Instead of going after Fav. outcomes, go after unfav. outcomes as its easier to calculate and use the formula in bold.

Total cases = any no. on all 3 dices = 6 x 6 x 6 = 216
Unfav. cases = NO 6's = 5 x 5 x 5 = 125

Fav. Cases = 216-125 = 91.



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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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