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MCQEasyJEE 2023Combinations (C(n,r))

JEE Mathematics 2023 Question with Solution

Let the number of elements in sets AA and BB be five and two respectively. Then the number of subsets of A×BA \times B each having at least 33 and at most 66 elements is:

  • A

    752752

  • B

    772772

  • C

    782782

  • D

    792792

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The number of elements in sets AA and BB are 55 and 22 respectively.

Find: The number of subsets of A×BA \times B having at least 33 and at most 66 elements.

First, find the number of elements in the Cartesian product:

n(A×B)=n(A)×n(B)=5×2=10n(A \times B) = n(A) \times n(B) = 5 \times 2 = 10

So, A×BA \times B has 1010 elements.

A subset with exactly rr elements can be chosen in (10r)\binom{10}{r} ways. Therefore, the required number of subsets is

(103)+(104)+(105)+(106)\binom{10}{3} + \binom{10}{4} + \binom{10}{5} + \binom{10}{6}

Now compute each term:

(103)=120,(104)=210,(105)=252,(106)=210\binom{10}{3} = 120, \quad \binom{10}{4} = 210, \quad \binom{10}{5} = 252, \quad \binom{10}{6} = 210

Adding these values:

120+210+252+210=792120 + 210 + 252 + 210 = 792

Therefore, the number of subsets is 792792. The correct option is D.

Using symmetry of combinations

Given: n(A)=5n(A) = 5 and n(B)=2n(B) = 2.

Find: The number of subsets of A×BA \times B with size from 33 to 66.

Since

n(A×B)=10n(A \times B) = 10

the required count is

(103)+(104)+(105)+(106)\binom{10}{3} + \binom{10}{4} + \binom{10}{5} + \binom{10}{6}

Use the symmetry property

(106)=(104)\binom{10}{6} = \binom{10}{4}

So the sum becomes

(103)+2(104)+(105)\binom{10}{3} + 2\binom{10}{4} + \binom{10}{5}

That is,

120+2(210)+252=120+420+252=792120 + 2(210) + 252 = 120 + 420 + 252 = 792

Therefore, the required number of subsets is 792792.

Common mistakes

  • Using 5+2=75 + 2 = 7 instead of 5×2=105 \times 2 = 10 for the number of elements in A×BA \times B. This is wrong because a Cartesian product counts ordered pairs, so multiply the number of choices. Use n(A×B)=n(A)n(B)n(A \times B) = n(A)n(B).

  • Counting all subsets as the answer. This is wrong because 210=10242^{10} = 1024 gives every subset of A×BA \times B, while the question asks only for subsets having between 33 and 66 elements. Restrict the count to (103)+(104)+(105)+(106)\binom{10}{3} + \binom{10}{4} + \binom{10}{5} + \binom{10}{6}.

  • Forgetting one of the required sizes, especially omitting subsets with exactly 66 elements or including subsets with 22 or 77 elements. This is wrong because 'at least 33 and at most 66' includes only sizes 3,4,5,63,4,5,6. List the allowed sizes before applying combinations.

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