Mathematics - Permutation & Combination Question with Solution | TestHub

MathematicsPermutation & CombinationArrangement under ConstraintHard2 minPYQ_2022
MathematicsHardnumerical

Let b1b2b3b4 be a 4-element permutation with bi 1,2,3,,100 for 1i4 and bibj for ij, such that either b1, b2, b3 are consecutive integers or b2,b3,b4 are consecutive integers. Then the number of such permutations b1b2b3b4 is equal to ______.

Answer:
18915.00
Solution:

bi1,2,3100

Let A= set when b1 b2 b3 are consecutive

nA=97+97++97 added 98 times=97×98=9506

Similarly when b2 b3 b4 are consecutive

nB=97×98=9506

Now when b1 b2 b3 b4 are consecutive then nAB=97

nAUB=nA+nB-nAB

Number of required permutation =9506+9506-97=18915

Stream:JEESubject:MathematicsTopic:Permutation & CombinationSubtopic:Arrangement under Constraint
2mℹ️ Source: PYQ_2022

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