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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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ans is 515
9 1------- 10C9*9C1
7 3-------10C7*9C3
5 5--------10C5*9C5
3 7
1 9 add all
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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*5C5
Number of ways to select 7 fruit for 1st child & 3 fruits to second child is :10C7*3C3
Number of ways to select 9 fruit for 1st child & 1 fruits to second child is :10C9*1C1
Total = \(10C1*9C9 + 10C3*7C7 + 10C5*5C5 + 10C7*3C3 + 10C9*1C1\\
=10 + 10*9*8/3*2. + 10*9*8*7*6/ (5*4*3*2). + 10*9*8/(3*2) +10\\
=10+120+ 2*9*7*2+ 10*3*4 +10\\
=10+120+36*7+120+10\\
=10+120+252+120+10\\
=260+252=512\)

Hence Option C is the answer
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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*5C5
Number of ways to select 7 fruit for 1st child & 3 fruits to second child is :10C7*3C3
Number of ways to select 9 fruit for 1st child & 1 fruits to second child is :10C9*1C1
Total =
\(10C1*9C9 + 10C3*7C7 + 10C5*5C5 + 10C7*3C3 + 10C9*1C1\)
\(=10 + 10*9*8/3*2. + 10*9*8*7*6/ (5*4*3*2). + 10*9*8/(3*2) +10\)
\(=10+120+ 2*9*7*2+ 10*3*4 +10\)
\(=10+120+36*7+120+10\)
\(=10+120+252+120+10\)
\(=260+252=512\)

Hence Option C is the answer
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(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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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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Total 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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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
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sonusaini1

Total 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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