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Olympiad workbook
Exam
Olympiad workbook
Subject
IOQM
Chapter
Permutation and Combination

Olympiad workbook · IOQM · Permutation and Combination

For any finite non empty set X of integers, let (X) denote the...

For any finite non empty set XX of integers, let max⁡(X)\max (X) denote the largest element of XX and ∣X∣|X| denote the number of elements in XX. If NN is the number of ordered pairs ( A,BA, B ) of finite non-empty sets of positive integers, such that max⁡(A)×∣B∣=12; and ∣A∣×max⁡(B)=11\begin{aligned} & \max (A) \times|B|=12 ; \text { and } |A| \times \max (B)=11 \end{aligned} and NN can be written as 100a+b100 a+b where a,ba, b are positive integers less than 100 , find a+ba+b

Correct answer

43

Step-by-step solution

\begin{aligned} & A=\left\{a_1, a_2, a_3 \ldots \ldots a_p\right\} \\ & B=\left\{b_1, b_2, b_3 \ldots . b_q\right\} \\ & \qquad \begin{aligned} & a_p \cdot q=12 \\ & p \cdot b_q=11\end{aligned} \\ & \text { Case-A : } p=11, b_q=1 \\ & A=\left\{a_1, a_2, a_3 \ldots . a_{11}\right\}, B=\{1\} \\ & \Rightarrow \quad a_{11}=12, \quad q=1 \\ & \therefore { }^{11} C_{10}= total \ ways \\ & \text { Case-B : } p=1, b_q=11 \\ & \text { (1) } A=\{12\}, B=\{11\} \rightarrow 1 \text { way } \\ & \text { (2) } A=\{6\}, B=\left\{b_1, 11\right\} \rightarrow{ }^{10} C_1 \text { ways } \\ & \text { (3) } A=\{4\}, B=\left\{b_1, b_2, 11\right\} \rightarrow{ }^{10} C_2 \text { ways } \\ & \text { (4) } A=\{3\}, B=\left\{b_1, b_2, b_3, 11\right\} \rightarrow{ }^{10} C_3 \text { ways } \\ & \text { (5) } A=\{2\}, B=\left\{b_1, b_2, b_3, b_4, b_5, 11\right\} \rightarrow{ }^{10} C_5 \text { ways } \\ & \text { (6) } A=\{1\}, B=\left\{b_1, b_2, \ldots . b_{11}, 11\right\} \rightarrow 0 \text { ways } \\ & \therefore \text { Total ways }=11+1+10+45+120+252 \\ & =439 \\ & =100 \times 4+39\end{aligned}$
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