How many distinct 9-digit numbers can be form | vedclass

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

Question

(C) The given number is $223355888$. The digits are: $2, 2, 3, 3, 5, 5, 8, 8, 8$.Total digits = $9$.Odd digits are: $3, 3, 5, 5$ (Total $4$ digits).Ev

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) The given number is $223355888$. The digits are: $2, 2, 3, 3, 5, 5, 8, 8, 8$.Total digits = $9$.Odd digits are: $3, 3, 5, 5$ (Total $4$ digits).Even digits are: $2, 2, 8, 8, 8$ (Total $5$ digits).Positions are: $1, 2, 3, 4, 5, 6, 7, 8, 9$.Even positions are: $2, 4, 6, 8$ (Total $4$ positions).Odd positions are: $1, 3, 5, 7, 9$ (Total $5$ positions).We need to place $4$ odd digits in $4$ even positions and $5$ even digits in $5$ odd positions.Number of ways to arrange $4$ odd digits $(3, 3, 5, 5)$ in $4$ positions = $\frac{4!}{2!2!} = \frac{24}{4} = 6$.Number of ways to arrange $5$ even digits $(2, 2, 8, 8, 8)$ in $5$ positions = $\frac{5!}{2!3!} = \frac{120}{2 	imes 6} = 10$.Total number of ways = $6 	imes 10 = 60$.

How many distinct $9$-digit numbers can be formed by rearranging the digits of the number $223355888$ such that all odd digits occupy even positions?

Similar Questions

Let $n$ be a number of digits. Using the three digits $2, 5, 7$,at least $900$ distinct $n$-digit numbers can be formed. What is the minimum value of $n$?

The number of ways to form a $7$-digit number using the digits $1, 2, 3, 4, 3, 2, 1$ such that odd digits always occupy odd places is:

${ }^{15} P_8 = A + 8 \cdot { }^{14} P_7 \Rightarrow A = $

How many numbers can be formed using the digits $1, 2, 3, 4, 3, 2, 1$ such that odd digits always occupy odd positions?

The number of integers greater than $3000$ that can be formed by using any number of digits from the set $\{0, 1, 2, 3, 4, 5\}$ without repetition is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module