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Permutations and Combinations question

2023 · 8 Apr · Shift 1 · Q34
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  5. /2023 · 8 Apr · Shift 1 · Q34

Permutations and Combinations question

2023 · 8 Apr · Shift 1 · Q34

JEE MainMathematicsPermutations and CombinationsMCQ+4 / −1
Let the number of elements in sets AAA and BBB be five and two respectively. Then the number of subsets of A×BA \times BA×B each having at least 3 and at most 6 elements is :
  1. A
    782
  2. B
    772
  3. C
    752
  4. D
    792
View written solutionFree

Correct answer: D

  1. Find the size of A×BA \times BA×B

Given:

  • ∣A∣=5|A| = 5∣A∣=5
  • ∣B∣=2|B| = 2∣B∣=2

So,

∣A×B∣=∣A∣⋅∣B∣=5⋅2=10|A \times B| = |A|\cdot |B| = 5\cdot 2 = 10∣A×B∣=∣A∣⋅∣B∣=5⋅2=10

Thus, A×BA \times BA×B has 101010 elements.

  1. Interpret the question

We need the number of subsets of a 101010-element set having at least 3 and at most 6 elements.

So required number is:

(103)+(104)+(105)+(106)\binom{10}{3}+\binom{10}{4}+\binom{10}{5}+\binom{10}{6}(310​)+(410​)+(510​)+(610​)
  1. Compute each binomial coefficient
(103)=120\binom{10}{3} = 120(310​)=120 (104)=210\binom{10}{4} = 210(410​)=210 (105)=252\binom{10}{5} = 252(510​)=252 (106)=(104)=210\binom{10}{6} = \binom{10}{4} = 210(610​)=(410​)=210
  1. Add them
120+210+252+210=792120+210+252+210 = 792120+210+252+210=792
  1. Match with options

The correct option is:

792\boxed{792}792​

which is Option D.

  1. Compare with stored correct answer

Stored correct answer: DDD

Our derived answer is also DDD.

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