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In how many ways 10 Oranges be distributed among 2 children if both mu
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03 Nov 2018, 10:18
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In how many ways 10 Oranges be distributed among 2 children if both must get odd no of Oranges? A. 513 B. 540 C. 512 D. 515 E. 516 Weekly Quant Quiz #7 Question No 9
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Re: In how many ways 10 Oranges be distributed among 2 children if both mu
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03 Nov 2018, 10:54
In how many ways 10 Oranges be distributed among 2 children if both must get odd no of apples? A. 513 B. 540 C. 512 D. 515 E. 516
lets say (1,9) or (3,7) or (5,5) , (7,3) and (9,1)
so 10C1*9C9 +10C3*7C7+ 10C5*5C5+10C7*3C3+10C9*1C1 =10 + 10*9*8/3*2. + 10*9*8*7*6/ (5*4*3*2). + 10*9*8/(3*2) +10 =10+120+ 2*9*7*2+ 10*3*4 +10 =10+120+36*7+120+10 =10+120+252+120+10 =260+252=512
Option C is the answer



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Re: In how many ways 10 Oranges be distributed among 2 children if both mu
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03 Nov 2018, 10:58
In how many ways 10 Oranges be distributed among 2 children if both must get odd no of apples? A. 513 B. 540 C. 512 D. 515 E. 516
The children must get odd number of apples.
So If A gets 1 B can Get (3,5,7,9) Similarly we get 4 cases ie 4^4
since there are 2 children we can multiply it by 2.
So ans:256*2: 512 C



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Re: In how many ways 10 Oranges be distributed among 2 children if both mu
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03 Nov 2018, 12:58
ans is 515 9 1 10C9*9C1 7 310C7*9C3 5 510C5*9C5 3 7 1 9 add all



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Re: In how many ways 10 Oranges be distributed among 2 children if both mu
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03 Nov 2018, 21:47
Only possible number of fruits that may be given is 1,3,5,7,9 So Number of ways to select 1 fruit for 1st child & 9 fruits to second child is : 10C1*9C9 Number of ways to select 3 fruit for 1st child & 7 fruits to second child is : 10C3*7C7 Number of ways to select 5 fruit for 1st child & 5 fruits to second child is : 10C5*5C5Number of ways to select 7 fruit for 1st child & 3 fruits to second child is : 10C7*3C3Number of ways to select 9 fruit for 1st child & 1 fruits to second child is : 10C9*1C1Total = \(10C1*9C9 + 10C3*7C7 + 10C5*5C5 + 10C7*3C3 + 10C9*1C1 =10 + 10*9*8/3*2. + 10*9*8*7*6/ (5*4*3*2). + 10*9*8/(3*2) +10 =10+120+ 2*9*7*2+ 10*3*4 +10 =10+120+36*7+120+10 =10+120+252+120+10 =260+252=512\) Hence Option C is the answer
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Re: In how many ways 10 Oranges be distributed among 2 children if both mu
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03 Nov 2018, 21:51
Only possible number of fruits that may be given is 1,3,5,7,9 So Number of ways to select 1 fruit for 1st child & 9 fruits to second child is : 10C1*9C9 Number of ways to select 3 fruit for 1st child & 7 fruits to second child is : 10C3*7C7 Number of ways to select 5 fruit for 1st child & 5 fruits to second child is : 10C5*5C5Number of ways to select 7 fruit for 1st child & 3 fruits to second child is : 10C7*3C3Number of ways to select 9 fruit for 1st child & 1 fruits to second child is : 10C9*1C1Total = \(10C1*9C9 + 10C3*7C7 + 10C5*5C5 + 10C7*3C3 + 10C9*1C1\) \(=10 + 10*9*8/3*2. + 10*9*8*7*6/ (5*4*3*2). + 10*9*8/(3*2) +10\) \(=10+120+ 2*9*7*2+ 10*3*4 +10\) \(=10+120+36*7+120+10\) \(=10+120+252+120+10\) \(=260+252=512\) Hence Option C is the answer
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Re: In how many ways 10 Oranges be distributed among 2 children if both mu
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04 Nov 2018, 03:00
(1,9) > 2 ways > 2* 10! / 1!9! = 20 (3,7) > 2 ways > 2* 10!/ 3!7! = 240 (5,5) > 1 way > 10!/5!5! = 252
Total 512



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Re: In how many ways 10 Oranges be distributed among 2 children if both mu
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05 Nov 2018, 01:20
Hi guys Is there any other way we can approach this problem. Thanks Sent from my Lenovo K33a42 using GMAT Club Forum mobile app
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In how many ways 10 Oranges be distributed among 2 children if both mu
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05 Nov 2018, 02:08
sonusaini1Total no of ways in which the 10 Oranges be given to 2 person = 2*2*2*…2 = 2^10 Now, there can only be two possibilities, either both get odd no of oranges or equal no of oranges. ( As the total no of oranges = even: Even = odd+odd or even +even) Also the probability of both cases are equal = ½ Hence, the number of ways in which both received odd no of oranges = ½ * (Total no of ways ) = ½ * 2^10 = 2^9 = 512
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Re: In how many ways 10 Oranges be distributed among 2 children if both mu
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05 Nov 2018, 06:00
gmatbusters wrote: sonusaini1Total no of ways in which the 10 Oranges be given to 2 person = 2*2*2*…2 = \(2^10\) Now, there can only be two possibilities, either both get odd no of oranges or equal no of oranges. ( As the total no of oranges = even: Even = odd+odd or even +even) Also the probability of both cases are equal = ½ Hence, the number of ways in which both received odd no of oranges = ½ * (Total no of ways ) = ½ *\(2^10\) = \(2^9\) = 512 Total no of ways in which the 10 Oranges be given to 2 person = 2*2*2*…2 = \(2^10\) Can you explain this? Also, shouldn't the question also mention that the oranges are distinct. It could easily be inferred as similar oranges. IN this case, the answer would have been vastly different.



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Re: In how many ways 10 Oranges be distributed among 2 children if both mu
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05 Nov 2018, 07:28
Response to query 1: Total no of ways in which the 10 Oranges be given to 2 person First orange can be given to any of the two person, no of ways = 2 Second orange can be given to any of the two person, no of ways = 2 . . . .Tenth orange can be given to any of the two person, no of ways = 2 total no of ways = = 2*2*2*…2 = 2^10 Response to query 2: Unless it is specifically mentioned that the tems are identical, they are taken as distinct.nitesh50 wrote: gmatbusters wrote: sonusaini1Total no of ways in which the 10 Oranges be given to 2 person = 2*2*2*…2 = \(2^10\) Now, there can only be two possibilities, either both get odd no of oranges or equal no of oranges. ( As the total no of oranges = even: Even = odd+odd or even +even) Also the probability of both cases are equal = ½ Hence, the number of ways in which both received odd no of oranges = ½ * (Total no of ways ) = ½ *\(2^10\) = \(2^9\) = 512 Total no of ways in which the 10 Oranges be given to 2 person = 2*2*2*…2 = \(2^10\) Can you explain this? Also, shouldn't the question also mention that the oranges are distinct. It could easily be inferred as similar oranges. IN this case, the answer would have been vastly different.
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Re: In how many ways 10 Oranges be distributed among 2 children if both mu
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