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In how many ways 10 Oranges be distributed among 2 children if both mu  [#permalink]

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7 00:00

Difficulty:   45% (medium)

Question Stats: 64% (02:30) correct 36% (03:15) wrong based on 61 sessions

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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  [#permalink]

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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

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
Current Student S
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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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Re: In how many ways 10 Oranges be distributed among 2 children if both mu  [#permalink]

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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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Re: In how many ways 10 Oranges be distributed among 2 children if both mu  [#permalink]

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1
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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Re: In how many ways 10 Oranges be distributed among 2 children if both mu  [#permalink]

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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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Manager  G
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Re: In how many ways 10 Oranges be distributed among 2 children if both mu  [#permalink]

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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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Re: In how many ways 10 Oranges be distributed among 2 children if both mu  [#permalink]

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Hi guys

Is there any other way we can approach this problem.

Thanks

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In how many ways 10 Oranges be distributed among 2 children if both mu  [#permalink]

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1
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
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gmatbusters wrote:
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.
Retired Moderator V
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Re: In how many ways 10 Oranges be distributed among 2 children if both mu  [#permalink]

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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 wrote:
gmatbusters wrote:
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.

_________________ Re: In how many ways 10 Oranges be distributed among 2 children if both mu   [#permalink] 05 Nov 2018, 07:28
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