How many 3-digit odd numbers can be formed from the digits 1,2,3,4,5,6 if:
 Number of places [(x), (y) and (z)] for them = 3Repetition is allowed and the 3-digit numbers formed are oddNumber of ways in which box (x) can be filled = 3 (by 1, 3 or 5 as the numbers formed are to be odd) m = 3 Number of ways of filling box (y) = 6 (∴ Repetition is allowed)n = 6 Number of ways of filling box (z) = 6 (∵ Repetition is allowed)p = 6 ∴ Total number of 3-digit odd numbers formed = m x n x p = 3 x 6 x 6 = 108(b) Number of ways of filling box (x) = 3 (only odd numbers are to be in this box )m = 3 Number of ways of filling box (y) = 5 (∵ Repetition is not allowed)n = 5 Number of ways of filling box (z) = 4 (∵ Repetition is not allowed)p = 4 ∴ Total number of 3-digit odd numbers formed= m x n x p = 3 x 5 x 4 = 60.
Math Expert Joined: 02 Sep 2009 Posts: 87082
In how many ways can 3 prizes be distributed among 5 students when [#permalink]
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Question Stats: Hide Show timer StatisticsIn how many ways can 3 different prizes be distributed among 5 students when no student gets all the prizes?(A) 60(B) 100(C) 120(D) 124(E) 125 M37-52 _________________
Math Expert Joined: 02 Sep 2009 Posts: 87082
Re: In how many ways can 3 prizes be distributed among 5 students when [#permalink]
Official Solution: In how many ways can 3 different prizes be distributed among 5 students when no student gets all the prizes? A. \(60\)B. \(100\)C. \(120\)D. \(124\)E. \(125\)Each of the three prizes can be assigned to any of the five students, so each prize has 5 options. So, the total number of ways to distribute 3 prizes among 5 students without the restriction is \(= 5*5*5=125\). This number contains 5 cases when each of the students gets all 3 prizes, so to get the desired number we should subtract these cases from 120. Thus, the final answer is \(125-5=120\). Answer: C _________________
Math Expert Joined: 02 Sep 2009 Posts: 87082
Re: In how many ways can 3 prizes be distributed among 5 students when [#permalink]
Math Expert Joined: 02 Sep 2009 Posts: 87082
Re: In how many ways can 3 prizes be distributed among 5 students when [#permalink]
Bunuel wrote: In how many ways can 3 different prizes be distributed among 5 students when no student gets all the prizes?(A) 60(B) 100(C) 120(D) 124 (E) 125 __________________________Anyone want to try this one? _________________
Manager Joined: 26 Nov 2020 Posts: 119 Location: India GPA: 3.9
Re: In how many ways can 3 prizes be distributed among 5 students when [#permalink] IMO Answer = 120
Re: In how many ways can 3 prizes be distributed among 5 students when [#permalink]
Bunuel wrote: In how many ways can 3 different prizes be distributed among 5 students when no student gets all the prizes?(A) 60(B) 100(C) 120(D) 124 (E) 125 Prices: 5 x 5 x 4 = 100IMO B _________________
Intern Joined: 09 Aug 2019 Posts: 5
Re: In how many ways can 3 prizes be distributed among 5 students when [#permalink] I think the answer should be 120. (5p3 + 3c2*5p2)5*5*4 cannot be the answer as I believe this situation is not the only situation (for eg. for the first prize lets assume out 5 students, student 1 got it. The second price now is also open to 5 students, lets say anyone apart from student 1 got it. Now the last prize is also open again to 5 students and not 4).Hope anyone would like to add my explanation. Posted from my mobile device
Manager Joined: 26 Nov 2020 Posts: 119 Location: India GPA: 3.9
Re: In how many ways can 3 prizes be distributed among 5 students when [#permalink]
kshv wrote: I think the answer should be 120. (5p3 + 3c2*5p2)5*5*4 cannot be the answer as I believe this situation is not the only situation (for eg. for the first prize let's assume out of 5 students, student 1 got it. The second price now is also open to 5 students, let's say anyone apart from student 1 got it. Now the last prize is also open again to 5 students and not 4).Hope anyone would like to add my explanation. Posted from my mobile device Yes. Understood my mistake. Your answer makes sense. It can also be solved rather quickly by subtracting the number of cases when all gifts go to 1 student from the total number of cases.
Intern Joined: 04 Mar 2021 Posts: 4
Re: In how many ways can 3 prizes be distributed among 5 students when [#permalink] All possible distributions: 5x5x5 = 125 (Including students get all three prices)All distributions where students get all prizes: 5 (Since there are only 3 prizes, only one student can get all three at the same time, since we have 5 students it's 5.)125-5 = 120 IMO C
Re: In how many ways can 3 prizes be distributed among 5 students when [#permalink] Since all 3 prizes (different) cannot be given to one student the distribution will be in the following ways: 2-1-0-0-0 First we will select one student who gets 2 prizes = 5c1 ways, then we will select which 2 prizes he gets = 3c2 ways. And then we will be left with 4 students and one prize, which will be distributed in 4c1 ways.Total = (5c1*3c2)*(4c1) = 60 ways 1-1-1-0-0 First we will select three students who gets 1 prize each = 5c3 ways, then we will select which prize each student gets = 3c1*2c1*1c1 ways. Total = 5c3*3c1*2c1*1c1 = 60 waysWe can also select three students first and then permute 3 prizes among them the answer will be same = 5c3*3p3Total ways = 60+60= 120 Option A
VP Joined: 10 Jul 2019 Posts: 1422
Re: In how many ways can 3 prizes be distributed among 5 students when [#permalink] Let the 3 distinct prizes be: X —-Y—-ZLet the 5 people be: A, B, C, D, and ESince no one person can receive all 3 prizes, there are 2 ways we can distribute the prizes:[1–1–1] or [2–1](Scenario 1)Distribute the prizes so that 3 different people each get 1 prize (1st)out of the 5 people, how many ways are there to choose the 3 people who will get the prize:“5 choose 3” = 5! / 2! 3! = 10And(2nd)for each of the 10 possible groups, we can divide the 3 distinct prizes among 3 different people in:3! = 6 waysScenario 1:(10) * (6) = 60 waysScenario 2:Distribute the prizes so that 1 person gets 2 of the distinct prizes and another 2nd person gets 1 of the prizes [2–1](1st) how many ways are there to choose the 2 people out of the 5 total who will receive the prizes?“5 choose 2” = 5! / 2! 3! = 10 waysAnd(2nd) how many ways are there to divide the 3 prizes up such that there is one stack of 2 prizes and another stack of 1 prize“3 choose 2” * “1 choose 1” = 3 waysAnd(3rd)how many ways are there to distribute the two distinct “stacks” of prizes to the 2 people chosen?2! = 2 waysScenario 2:(10) * (3) * (2) = 60 waysAnswer60 + 60 =120 ways Posted from my mobile device
Re: In how many ways can 3 prizes be distributed among 5 students when [#permalink] Total no of options : - 5 x 5 x 5 = 125 Less ways in which all prices receiving by single student: 5 Answer is 125 - 5 = 120
Manager Joined: 06 Apr 2021 Posts: 74
Re: In how many ways can 3 prizes be distributed among 5 students when [#permalink] In case you were not quick to pick the "Total Possibilities-Possibilities of one having all 3 trophies" methodYou could use the permutation method I used:Poss. of 3 out of 5 having a trophy each: 5*4*3=60Poss. of 2 out of 5 having the trophies: 5*4=20, Since there are 3 different trophies they can be distributed in 3 ways (3C2)So 20*3=60 Add both scenarios= 60+60=120 (C) By far the best method is 125-5=120 but it didn't click to me as I started working on this. Dammit
Re: In how many ways can 3 prizes be distributed among 5 students when [#permalink] 26 May 2021, 01:38 |
In how many ways can 3 prizes be given to 5 students
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