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Find the number of ways to choose a pair $\{a,b\}$ of distinct numbers from the set $\{1,2,...,50\}$ such that

i) $|a-b| = 5$;

ii) $|a-b| \leq 5 $

My thoughts:

For (i) For every 6 consecutive numbers, there's a pair. But I can't do $C_{6}^{50}$ as it's not every 6 number combinations I want.

Then answer given is (i)45 (ii)235

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4 Answers 4

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For the first problem, the leftmost (smallest) of the chosen numbers can be any of $1,2,3,\dots,45$. Then the other number is determined.

A similar idea will take care of the second problem. The number of choices is $49+48+47+46+45$.

The same idea works in general. For the generalization of the second problem, we are summing a simple arithmetic sequence.

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  • $\begingroup$ I don't understand how to solve the second problem. Why do we have to sum up? Thanks! $\endgroup$ Sep 29, 2014 at 18:32
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    $\begingroup$ If the difference is $1$, the leftmost can be chosen in $49$ ways. If the difference is $2$, the leftmost can be chosen in $48$ ways, and so on to $45$ ways. Absolute value of difference $\le 5$ breaks down into $5$ cases, absolute value is $1$, absolute value is $2$, is $3$, is $4$, is $5$. $\endgroup$ Sep 29, 2014 at 18:35
  • $\begingroup$ I'm wondering why $45 \times n_{2}^{6}$ doesn't give the same answer? $\endgroup$ Sep 29, 2014 at 19:57
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    $\begingroup$ I see no reason that it should, since I do not see $45\times \binom{6}{2}$ as counting what we want to count. $\endgroup$ Sep 29, 2014 at 20:09
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For the first one Consider number (1, 6) and (2,7) .....(45, 50) there are 45 ways

For the second one consider the number 1 and 6 again and the number between them. If we select any two numbers from these the difference is less than 5 and the number of ways we can select any two numbers from these is C(6, 2). We have more numbers

2, 3, 4, 5, 6, 7

Now here since we have already chosen combinations with the numbers 2, 3, 4, 5, 6 we have to choose combinations with the number 7 so choosing these pair's with 7 is 5 ways. Now considering the same argument with other numbers gives the total number of ways is equal to C(6, 2)+44*5=235

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  • $\begingroup$ Sorry, just realized the answer to the second question is wrong. The answer is supposed to be 235. $\endgroup$ Sep 29, 2014 at 18:30
  • $\begingroup$ Edited the answer @Delvacode .... $\endgroup$
    – Jasser
    Sep 30, 2014 at 4:50
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For i) it's all the integers that lie on the lines $y=x+5$ and $y=x-5$ in the square $[1,50]^2$. These are symmetric so you need only count one of them and multiply by 2.

$$2\sum_{x=1}^{45} 1 = 90$$

For ii) You can count all that above or on $y=x+5$ as $x$ moves from 1 to 45, multiply by 2, and subtract from 2500.

$$2500-2\sum_{x=1}^{45}(45-x)=520$$

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For the second problem, you can consider the two cases where the smaller number a satisfies

1) $1\le a\le 45$ and 2) $46\le a\le 49$.

The first case gives $45\times5$ possibilities, and the second case gives $4+3+2+1=10$ possibilities, for a total of 235 choices.


Alternatively, you can let $x_1,x_2,x_3$ be the number of integers in the 3 gaps created by a and b.

Then $x_1+x_2+x_3=48$ where $x_1\ge0, \;0\le x_2\le 4, \;x_3\ge0$;

so the number of solutions is given by $\binom{50}{2}-\binom{45}{2}=235.$

(I am assuming the pairs are unordered.)

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