All the five digits numbers in which each suc | vedclass

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Question

All the possible number are $^{9} \mathrm{C}_{5}$ (none containing the digit $0$ $)=126$

Total numbers starting with

$1 = {\,^8}{{\rm{C}}_4} = 7

Vedclass · Accepted Answer

$5$

Vedclass · Answer

All the possible number are $^{9} \mathrm{C}_{5}$ (none containing the digit $0$ $)=126$

Total numbers starting with

$1 = {\,^8}{{m{C}}_4} = 70$

${ 	ext { (using }2,3,4,5,6,7,8,9)} $

${	ext { Total starting with }}$

$\qquad 23 = {\,^6}{{m{C}}_3} = 20$

$(	ext { using } 4,5,6,7,8,9)$

${	ext { Total numbers starting with }} $

$\qquad 245 = {\,^4}{{m{C}}_2} = 6$

${m{ 9}}{{m{7}}^{th}}{m{  number }} = $

All the five digits numbers in which each successive digit exceeds its predecessor are arranged in the increasing order of their magnitude. The $97^{th}$ number in the list does not contain the digit:

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