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Olympiad workbook
Exam
Olympiad workbook
Subject
IOQM
Chapter
Permutation and Combination

Olympiad workbook · IOQM · Permutation and Combination

Let N be the number of ways of choosing a subset of 5 distinct...

Let NN be the number of ways of choosing a subset of 5 distinct numbers from the set {10a+b:1≤a≤5,1≤b≤5}\{10 a+b: 1 \leq a \leq 5,1 \leq b \leq 5\} where a,b\mathrm{a}, \mathrm{b} are integers, such that no two of the selected numbers have the same units digit and no two have the same tens digit. What is the remainder when NN is divided by 73 ?

Correct answer

47

Step-by-step solution

10a+b;1≤a≤5,1≤b≤510 a+b ; 1 \leq a \leq 5,1 \leq b \leq 5 Let us divide numbers into different sets, such as Set 1={11,12,13,14,15}1=\{11,12,13,14,15\} Set 2={21,22,23,24,25}2=\{21,22,23,24,25\} Set 3={31,32,33,34,35}3=\{31,32,33,34,35\} Set 4={41,42,43,44,45}4=\{41,42,43,44,45\} Set 5={51,52,53,54,55}5=\{51,52,53,54,55\} Now to make a number having no two digits and no two ten's digit same, we can select any 1 number from each of set 1 , set 2 , set 3 , set 4 , set 5 . ∴ No. of ways =5C1×4C1×3C1×2C1×1C1=120∴120÷73⇒ Remainder =47\begin{aligned} & \therefore \text { No. of ways }={ }^5 C_1 \times{ }^4 C_1 \times{ }^3 C_1 \times{ }^2 C_1 \times{ }^1 C_1=120 \\ & \therefore 120 \div 73 \\ & \Rightarrow \text { Remainder }=47 \end{aligned}
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