Question:
How many numbers with two or more digits can be formed with the digits 1,2,3,4,5,6,7,8,9, so that in every such number, each digit is used at most once and the digits appear in the ascending order?
Correct Answer: 502
As the digits appear in ascending order in the numbers, number of ways of forming a n-digit number using the 9 digits = $^{9} \mathrm{C}_{\mathrm{n}}$
Number of possible two-digit numbers which can be formed = $^{9} \mathrm{C}_{2}+^{9} \mathrm{C}_{3}+^{9} \mathrm{C}_{4}+^{9} \mathrm{C}_{5}+^{9} \mathrm{C}_{6}+^{9} \mathrm{C}_{7}+^{9} \mathrm{C}_{8}+^{9} \mathrm{C}_{9}$
$={{2}^{9}}-\left( ^{9}{{\text{C}}_{1}}{{+}^{9}}{{\text{C}}_{1}} \right)$
$=512-(1+9)=502$
As the digits appear in ascending order in the numbers, number of ways of forming a n-digit number using the 9 digits = $^{9} \mathrm{C}_{\mathrm{n}}$
Number of possible two-digit numbers which can be formed = $^{9} \mathrm{C}_{2}+^{9} \mathrm{C}_{3}+^{9} \mathrm{C}_{4}+^{9} \mathrm{C}_{5}+^{9} \mathrm{C}_{6}+^{9} \mathrm{C}_{7}+^{9} \mathrm{C}_{8}+^{9} \mathrm{C}_{9}$
$={{2}^{9}}-\left( ^{9}{{\text{C}}_{1}}{{+}^{9}}{{\text{C}}_{1}} \right)$
$=512-(1+9)=502$
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