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Let $\ x_1\ x_2\ x_3\ x_4\ x_5\ x_6$ be a six digit number, find the number of such numbers.

Case 1) $\ x_1 <\ x_2 <\ x_3 <\ x_4 <\ x_5<\ x_6$

Case 2) $\ x_1 <\ x_2 <\ x_3 $=$\ x_4 <\ x_5<\ x_6$

Case 3) $\ x_1 <\ x_2 <\ x_3 $ $\le$ $\ x_4 <\ x_5<\ x_6$

I have no idea how to approach this problem ,

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  • $\begingroup$ Least number is 123456 and the greatest is 456789 $\endgroup$ – Samar Imam Zaidi Oct 20 '17 at 11:01
  • $\begingroup$ Hint: For case $(1)$, try ${9\choose 6}$. Why does that help? $\endgroup$ – AnotherJohnDoe Oct 20 '17 at 11:07
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Hint:

Consider choosing $6$ distinct digits from $\{1,2,3,4,5,6,7,8,9\}$

  • case 1: The 6 chosen digits can be arranged in only one order. Hence the answer is $\binom{9}{6}$

  • case 2: we need to choose only $5$ digits as $2$ of $6$ are equal. So the answer is $\binom{9}{5}$

  • case 3: It is the union of the above cases.

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  • $\begingroup$ I want you to approve my understanding ,we select six number from 9 number. That is the combination part and now for increasing number among the selected number their is one way to arrange the nunber that is the permutation part. Now i understood the logic behind it $\endgroup$ – Samar Imam Zaidi Oct 20 '17 at 11:16
  • $\begingroup$ @Samar yeah that's exactly what we are doing here! $\endgroup$ – samjoe Oct 20 '17 at 11:18

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