JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Permutations and Combinations

The number of natural numbers less than 7,000 which can be formed by using the digits

0, 1, 3, 7, 9

(repetition of digits allowed) is equal to:

Answer: (B)

Solution

Count by the number of digits, drawing from

{0, 1, 3, 7, 9}

with repetition allowed. One-digit numbers: the four non-zero digits,

4

. Two-digit: first digit from the four non-zero, second from all five,

4 \times 5 = 20

. Three-digit:

4 \times 5 \times 5 = 100

. Four-digit but under

7000

: the lead digit must be

1

3

(two choices) with the other three free,

2 \times 5^{3} = 250

. Altogether

4 + 20 + 100 + 250 = 374

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Single correct option