Method-1

We need 4 siblings who must come from 4different sibling pairs therefore
Step-1: Select 4 sibling pairs = 8C4 ways = 70

Step-2: Select one child from each selected sibling pair = 2*2*2*2 = 16 ways

Total ways = 70*16 = 1120

Answer: Option C

Method-2

First sibling may be chosen in 16 ways
Second child may be chosen in 14 ways (leaving the selected child and his/her sibling)
Third child may be chosen in 12 ways (leaving the selected children and their siblings)
FOurth child may be chosen in 10 ways (leaving the selected children and their siblings)

Total Ways of making groups = 16*14*12*10

But this includes arrangement cause second child could come as first selection or vice versa so we need to exclude teh arrangements of 4 children from result

Required selection = 16*14*12*10 / 4! = 1120

Answer: Option C
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all are boys ; 8c4 = 70
all are girls ; 8c4 = 70
3 boys 1 girl ; 8c3*5c1 =56*5 ; 280
1 boy 3 girls ; 8c1*7c3 = 8*35 ; 280
2 boys 2 girls ; 8c2*6c2 = 420
total 70+70+280+280+420 ; 1120

OPTION C

A singing choir of 4 children is to be formed. The selection pool consists of 8 siblings pairs. In how many ways can a choir be formed if both siblings cannot be selected together?
(A) 360
(B) 720
(C) 1120
(D) 2400
(E) 26880

Check your mistakes

Method-3
case 1 ;
all are boys ; 8c4 = 70
all are girls ; 8c4 = 70
3 boys 1 girl ; 8c3*5c1 = 56*5 = 280 (Select girl such that she is not sibling of any chosen boy)
1 boy 3 girls ; 8c1*7c3 =8*35 = 280 (Select girls such that they are not sibling of chosen boy)
2 boys 2 girls ; 8c2*6c2 = 420

Total = 70+70+280+280+420 = 1120

Method-4
case 2 ;
total possible cases - ( set of siblings)

16c4- (8c2) - (8C1*(14*12/2)) = 1820 - 28 - (8*84) = 1820 - 28 - 672 = 1120

8C2 = selecting 2 sibling pairs out of 8 sibling pairs

8C1 = selecting 1 sibling pair out of 8 sibling pairs
(14*12/2) - Selecting 2 out of 14 such that they are not sibling pairs so 14 ways to choose 1st and 12 ways to choose second leaving chosen person and his cousin and then divided by 2 in order to exclude all arrangements of those two chose

Archit3110
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GMATinsight thank you .. understood where i went wrong...

i think the question should be 4 pair of children.

I used the same approach but couldn't figure why should the number be divided by 4!. There is no repetition right. Could you explain please

(1st) select which 4 of the 8 sibling pairs will donate 1 or the siblings to the team:
How many unique ways can we do this?

8 c 4 = 8! / (4! 4!) =

(8 * 7 * 6 * 5) / (4 * 3 * 2 * 1) =

(7 * 2 * 5) = 7 * 10 =

70 ways

AND

(2nd)for each of these 70 unique ways to pick out 4 of the sibling pairs, each sibling pair can donate the Brother OR Sister to the team:

(2) (2) (2) (2) = 16


Total ways = 70 * 16 =

1.120

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Can someone please explain why cant we do 16C4 - (8C1 * 14C2) ways?

I believe I can maybe help.

You are trying to take:

(Total Possible Outcomes)

-

(unfavorable outcomes)

=

Favorable outcomes


However:

(8 c 1) (14 c 2) ——-> will Overcount the unfavorable outcomes and you will end up removing some Unfavorable outcomes multiple times and the Answer will be too low


For instance:

Imagine the 16 people are broken up into couples as follows:

(A1 , A2) (B1 , B2) (C1 , C2) (D1 , D2)

(E1, E2) (F1 , F2) (G1 , G2) (H1 , H2)


Case 1: one of the teams you will end up counting as the removed teams is:

For 8 c 1 ———-> the 1 couple chosen is (A1 , A2)

and then

for 14 c 2 ———> you choose (B1 , B2)


However, you will also choose this same team again when:

Case 2:

For 8 c 1 ———> you choose couple (B1 , B2)

And then

For 14 c 2 ———> you choose (A1 , A2)


In each of those cases, the team has the same four people, but you will have counted that unique team multiple times

There are other such instances of over counting for the (# of Unfavorable Cases)

Does that help a little?

That’s why I always try to break these things out scenario by scenario. Even if it may take a little bit longer, it avoids over counting.





Posted from my mobile device

Why 4! is divided plz explain

Why 4! is divided plz explain[/quote]

Santanubarik, As the order of the selected children does not matter. Picking C1,C2, C3 and C4 is the same as picking C4,C3,C2 and C1.