The number of 4-digit integers in the closed | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) We need to find the number of $4$-digit integers in the interval $[2022, 4482]$ using the digits ${0, 2, 3, 4, 6, 7}$.Case $(1)$: Integers startin

Vedclass · Accepted Answer

$569$

Vedclass · Answer

(B) We need to find the number of $4$-digit integers in the interval $[2022, 4482]$ using the digits ${0, 2, 3, 4, 6, 7}$.Case $(1)$: Integers starting with $202...$Numbers are $2022, 2023, 2024, 2026, 2027$. Total = $5$.Case $(2)$: Integers starting with $203..., 204..., 206..., 207...$First digit is $2$,second is $0$,third can be ${3, 4, 6, 7}$ ($4$ choices),fourth can be any of ${0, 2, 3, 4, 6, 7}$ ($6$ choices).Total = $4 	imes 6 = 24$.Case $(3)$: Integers starting with $22..., 23..., 24..., 26..., 27...$First digit is $2$,second can be ${2, 3, 4, 6, 7}$ ($5$ choices),third can be any ($6$ choices),fourth can be any ($6$ choices).Total = $5 	imes 6 	imes 6 = 180$.Case $(4)$: Integers starting with $3...$First digit is $3$,second,third,and fourth can be any ($6$ choices each).Total = $6 	imes 6 	imes 6 = 216$.Case $(5)$: Integers starting with $40..., 42..., 43..., 44...$ up to $4482$.For $40..., 42..., 43...$: $3 	imes 6 	imes 6 = 108$.For $440..., 442..., 443...$: $3 	imes 6 = 18$.For $4440, 4442, 4443, 4444, 4446, 4447$: $6$ numbers.For $4460, 4462, 4463, 4464, 4466, 4467$: $6$ numbers.For $4470, 4472, 4473, 4474, 4476, 4477$: $6$ numbers.Sum for Case $(5) = 108 + 18 + 6 + 6 + 6 = 144$.Total = $5 + 24 + 180 + 216 + 144 = 569$.

The number of $4$-digit integers in the closed interval $[2022, 4482]$ formed by using the digits $0, 2, 3, 4, 6, 7$ is:

Similar Questions

In a group of $3$ girls and $4$ boys,there are two boys $B_1$ and $B_2$. The number of ways in which these girls and boys can stand in a queue such that all the girls stand together,all the boys stand together,but $B_1$ and $B_2$ are not adjacent to each other,is:

How many words can be formed by taking two identical and two distinct letters from the letters of the word $COMBINATION$?

What is the remainder when $1! + 2! + 3! + \dots + 200!$ is divided by $14$?

The number of positive integral solutions of $xyz = 60$ is

If the letters of the word $INTERMEDIATE$ are arranged,what is the probability that the two $E$'s do not come together?

Vedclass Test Series

Exam Paper Generator

Online Exam Module