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So, the question is- A rack has 5 different pairs of shoes. Number of ways in which 4 different shoes can be chosen so that no two shoes are from the same pair.

Now, what they have done is-

${5 \choose 4}$${2 \choose 1}$${2 \choose 1}$${2 \choose 1}$${2 \choose 1}$

Now , they said that they have taken ${5 \choose 4}$ to select any 4 pairs from the 5 and ${2 \choose 1}$ is for selecting any one shoe. But, pairs and shoes are different. So, how are they multiplying like this?

Kindly help me!

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    $\begingroup$ First you choose the 4 pairs that you will take shoes from, and then one shoe from each of those pairs. $\endgroup$ – Casteels Feb 21 '17 at 18:29
  • $\begingroup$ But, one is pairs and the other is shoes. They are different, right? I am unable to get the logic. $\endgroup$ – Abhishekstudent Feb 21 '17 at 18:34
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    $\begingroup$ Label the shoes $a_1,a_2,b_1,b_2,c_1,c_2,d_1,d_2,e_1,e_2$ where two shoes with the same letter belong to the same pair. First step: pick which letters are used. (there are five letters: a,b,c,d,e, and we are choosing four of them, so $\binom{5}{4}$ options here) Second step: for each letter chosen, pick the subscript number. Does that help? $\endgroup$ – JMoravitz Feb 21 '17 at 18:35
  • $\begingroup$ OHHH! I think that helps! $\endgroup$ – Abhishekstudent Feb 21 '17 at 18:43
  • $\begingroup$ this question assumes that two shoes from same pair are distinct. if considered identical, it may be more difficult. $\endgroup$ – Kiran Feb 21 '17 at 19:13
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You're trying to find the number of ways 4 shoes can be chosen so that no two are from the same pair.

If no two are from the same pair, all 4 shoes need to come from 4 different pairs, hence $\binom{5}{4}$. You're picking 4 of the 5 pairs of shoes to pick a single shoe from each.

Now you need to pick one shoe from each pair. So you go to one of the pairs and pick one of them, hence $\binom{2}{1}$. Then you do this for the next and the next until you've chosen a shoe from each pair. For each of the two choices in pair $p_1$ you make, there are two choices in $p_2$ you can make and so on.

Say you choose the left shoe from $p_1$ and the right shoe from $p_2$. That's one possible combination. Or you could have chosen the left shoe from $p_2$ as well. That's a total of two combinations. You have two more if you had chosen the right shoe from $p_1$. That's why we multiply.

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  • $\begingroup$ I got it! Thanks a ton! $\endgroup$ – Abhishekstudent Feb 22 '17 at 14:08
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Pairs are just groupings of shoes. So you're picking groups of shoes with the $\binom 54$ - because those are the restrictions on the choice of shoes - and then picking from within the groups to make the final shoe selection.

You could recast the problem as say picking four children from a group of ten children made up of two children from each of five families, with the restriction that you can't pick two children from the same family. Then you could say you pick four of five families, but what you're actually doing is picking four of the five one-pick-only groups of children; not really families. You could make similar groupings for other reasons (curly hair or whatever), but still the pick would be of subgroups of the final pick based on the restrictions of the problem.

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If you have a rack of 5 pairs, its 10 shoes. However, no two shoes can be a pair, and I assume no pairs are identical.

Then you have 5 shoes from which to choose 4 shoes. Since you are omitting all the left or right shoes, or a mix of both, you don't care about repetition, and you can then have any order of 5 shoes from which to choose 4. This is 5 ways as you just shift the unchosen slot through every shoe. For a bit of starting intuition, check for the following in the question: Ordered or unordered? With or without repetition?

And just adapt your question to fall in either of the 4 categories above

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