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JEE Main 202125 FebruaryMorning Shift
Exam
JEE Main
Year
2021
Date
25 February
Shift
Morning Shift
Subject
Mathematics
Chapter
Permutation and Combination

JEE Main 2021 · 25 February · Morning Shift · Mathematics · Permutation and Combination

The total number of numbers, lying between 100 and 1000 that can be...

The total number of numbers, lying between 100100 and 10001000 that can be formed with the digits 1,2,3,4,5,1, 2, 3, 4, 5, if the repetition of digits is not allowed and numbers are divisible by either 33 or 5,5, isThe total number of numbers, lying between 100 and 1000 that can be formed with the digits 1, 2, 3, 4, 5, if theThe total number of numbers, lying between 100 and 1000 that can be formed with the digits 1, 2, 3, 4, 5, if the

Correct answer

32

Step-by-step solution

We need three digits numbers.

Since 1+2+3+4+5=151 + 2 + 3 + 4 + 5 = 15

So, number of possible triplets for multiple of 1515 is 1×2×21 \times 2 \times 2

so =4×⌊3+4×3−1×2×⌊2=32= 4 \times ⌊ 3 + 4 \times 3 - 1 \times 2 \times ⌊ 2 = 32

Alternate:

The total number of numbers, lying between 100 and 1000 that can be formed with the digits 1, 2, 3, 4, 5, if the

divisible by 5=5 =4×3=124 \times 3 = 12

The total number of numbers, lying between 100 and 1000 that can be formed with the digits 1, 2, 3, 4, 5, if the

Divisible by 33

12→3,4,5→3!=615→2,3,4→3!=624→1,3,5→3!=642→1,2,3→3!=6\begin{array}{ll}12 \to 3, 4, 5 \to 3 ! = 6 \\ 15 \to 2, 3, 4 \to 3 ! = 6 \\ 24 \to 1, 3, 5 \to 3 ! = 6 \\ 42 \to 1, 2, 3 \to 3 ! = 6\end{array}24</p><p>Divisibleby</p><p>Divisible by15</p><p></p><p>1, 4, 5; 4, 1, 5; 3, 4, 5; 4, 3, 5</p><p>RequiredNo.</p><p>Required No.= 24 + 12 - 4 = 32$

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