Question 1133637
choose {{{4}}} out of {{{10}}}
 there can be ONLY {{{1}}} way the {{{4}}} can be arranged in ascending order, so selection is what we are looking for and NOT arrangements


{{{10C4=10!/(6!4!)=(10*9*8*7)/(4*3*2)=210}}}


BUT it contains numbers with {{{0}}} in thousands digit making these 3-digit numbers

so {{{9C3=9!/(6!3!)=(9*8*7)/(3*2)=84}}}.......deduct it from {{{210}}}


{{{210-84=126}}}

your ans: there is {{{126}}} four digit increasing numbers