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Different possible dog-boy pairs are 56 (4*14). but if the question says that how many dog-boy pairs are possible, when total number of boys are 4 and total number of dogs are 14, then the answer will be 4. i.e. equal to the number of boys.

why use perms in this case? maybe because if we have 4 boys to be paired, then we have to choose 4 dogs out of the 14 available....so 14P4....this is if the choiche of all the pairs is istantaneous...otherwise I think that simply 14*4 is good. anyway, what is OA? what is the right reasoning?

There are 4 boys to choose from and 14 dogs to choose from 4*14 = 56 possible pairs

This is the definition of a basic combo question. You guys are just so used to doing uber-high level math you tried to make it more complicated than it really is

EDIT: since they're being put into pairs of 2 you can write it 4C1*14C1 = 56.

Last edited by eschn3am on 09 Jan 2008, 07:04, edited 1 time in total.

There are 4 guys, 2 girls and 14 dogs. How many dog-boys pairs are possible.

This is a great problem because most of us are so use to complicated combinatorics questions we often miss that its simply

4*14.

Here is a question from a wonderlic test (test used to draft NFL players) that is similar to this problem:

Randolph has 8 ties, 6 pairs of pants, and 4 dress shirts. How many days could he possibly go without wearing the same combination of these three items.

On the wonderlic test you have about 15-30sec to answer this question.

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