The number of four-digit numbers that can be | vedclass

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

Question

(D) Step $1$: Choose $2$ distinct digits out of $9$ available digits ($1$ to $9$). This can be done in $^9C_2 = \frac{9 	imes 8}{2} = 36$ ways.Step $

Vedclass · Accepted Answer

$504$

Vedclass · Answer

(D) Step $1$: Choose $2$ distinct digits out of $9$ available digits ($1$ to $9$). This can be done in $^9C_2 = \frac{9 	imes 8}{2} = 36$ ways.Step $2$: For each pair of chosen digits,we need to form a $4$-digit number using both digits at least once.Step $3$: The total number of ways to fill $4$ positions using $2$ chosen digits is $2^4 = 16$.Step $4$: We must exclude the cases where only one digit is used (i.e.,all $4$ digits are the same). There are $2$ such cases (all $4$ digits are the first chosen digit or all $4$ digits are the second chosen digit).Step $5$: The number of valid $4$-digit numbers for each pair is $(2^4 - 2) = 14$.Step $6$: Total numbers = $36 	imes 14 = 504$.

The number of four-digit numbers that can be formed using all the digits from $1$ to $9$ (excluding zero) such that every number has exactly $2$ distinct digits is:

Similar Questions

What is the total number of ways of choosing $4$ cards from a pack of $52$ playing cards? In how many of these are the $4$ cards of the same colour?

$\binom{n}{r} \div \binom{n}{n-1} = \dots$

If $n$ is even and the value of $^nC_r$ is maximum,then $r = $

$A$ set contains $(2n + 1)$ elements. The number of subsets of the set which contain at most $n$ elements is:

The number of ways in which an examiner can assign $30$ marks to $8$ questions,giving not less than $2$ marks to any question,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module