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At the Chien Dog Show, there were 11 dogs in the semi-finals, of which 22 Dec 2016, 02:26
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At the Chien Dog Show, there were 11 dogs in the semi-finals, of which 4 will be selected for the finals. If two of the dogs selected will be Felix and Shadow, then how many groups of dogs are possible for the finals at the Chien Dog Show?

A. 36
B. 144
C. 720
D. 3600
E. 7920
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A
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At the Chien Dog Show, there were 11 dogs in the semi-finals, of which 22 Dec 2016, 02:46
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\(_9C_2 = 36\)

A
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shashank1tripathi
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At the Chien Dog Show, there were 11 dogs in the semi-finals, of which 22 Dec 2016, 04:43
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Selection of two dogs are already fixed. Then, to select rest of the 2 dogs = 9C2 = 9!/7!2! = 36.

Ans- A.
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At the Chien Dog Show, there were 11 dogs in the semi-finals, of which 26 Dec 2016, 01:44
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i think B should be the answer.

2x9x8 = 144
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At the Chien Dog Show, there were 11 dogs in the semi-finals, of which Updated on: 07 Jul 2017, 16:33
Originally posted by kimrani on 01 Jul 2017, 04:17.
Last edited by kimrani on 07 Jul 2017, 16:33, edited 1 time in total.
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Since the question asks "how many groups of dogs are possible for the finals at the Chien Dog Show", the order doesn't matter. Answer is 2C9 = 9!/(2!*7!)=36

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At the Chien Dog Show, there were 11 dogs in the semi-finals, of which 01 Jul 2017, 05:49
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This is combinations problem as the order of chosen will not matter.
Two dogs are already chosen so we have to choose 2 more dogs out of 9 available dogs so that can be done in 9C2 =36 ways
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At the Chien Dog Show, there were 11 dogs in the semi-finals, of which 01 Jul 2017, 06:40
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No. of ways of selecting 2 dogs out of 9 dogs = \(\frac{9*8}{1*2} = 36\). Ans - A.
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At the Chien Dog Show, there were 11 dogs in the semi-finals, of which 08 Dec 2018, 09:22
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why do we choose 2 dogs out of 9 for final? if there are 11 dogs in semifinal?
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At the Chien Dog Show, there were 11 dogs in the semi-finals, of which 15 Sep 2020, 12:15
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