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Functions question

2021 · 22 Jul · Shift 2 · Q38
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Functions question

2021 · 22 Jul · Shift 2 · Q38

JEE MainMathematicsFunctionsNumerical+4 / −1
Let A = {0, 1, 2, 3, 4, 5, 6, 7}. Then the number of bijective functions f : A →\to→ A such that f(1) + f(2) = 3 −-− f(3) is equal to
Numerical answer
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Correct answer: 720

  1. We need the number of bijections f:A→Af:A\to Af:A→A, where A={0,1,2,3,4,5,6,7},A=\{0,1,2,3,4,5,6,7\},A={0,1,2,3,4,5,6,7}, such that f(1)+f(2)=3−f(3).f(1)+f(2)=3-f(3).f(1)+f(2)=3−f(3).

Since fff is bijective, the values f(1),f(2),f(3)f(1),f(2),f(3)f(1),f(2),f(3) must be distinct elements of AAA.

  1. Rewrite the condition: f(1)+f(2)+f(3)=3.f(1)+f(2)+f(3)=3.f(1)+f(2)+f(3)=3. So we need to count the number of ordered triples (a,b,c)=(f(1),f(2),f(3))(a,b,c)=(f(1),f(2),f(3))(a,b,c)=(f(1),f(2),f(3)) with distinct entries from AAA such that a+b+c=3.a+b+c=3.a+b+c=3.

  2. Since all entries are nonnegative integers in AAA, the only way three distinct elements can sum to 333 is: 0+1+2=3.0+1+2=3.0+1+2=3. Other possibilities like 3+0+03+0+03+0+0 or 1+1+11+1+11+1+1 are not allowed because values must be distinct in a bijection.

Hence the set of values taken by (f(1),f(2),f(3))(f(1),f(2),f(3))(f(1),f(2),f(3)) must be exactly {0,1,2}\{0,1,2\}{0,1,2}.

  1. Count the number of ordered assignments of 0,1,20,1,20,1,2 to (f(1),f(2),f(3))(f(1),f(2),f(3))(f(1),f(2),f(3)): 3!=6.3!=6.3!=6.

  2. After fixing f(1),f(2),f(3)f(1),f(2),f(3)f(1),f(2),f(3), the remaining five elements of the domain, {0,4,5,6,7},\{0,4,5,6,7\},{0,4,5,6,7}, must map bijectively to the remaining five codomain values, {3,4,5,6,7}.\{3,4,5,6,7\}.{3,4,5,6,7}. The number of such bijections is 5!=120.5!=120.5!=120.

  3. Therefore total number of required bijections is 3!⋅5!=6⋅120=720.3!\cdot 5!=6\cdot 120=720.3!⋅5!=6⋅120=720.

  4. Comparison with stored answer: Derived answer = 720720720, stored correct answer = 720720720. They match.

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