Question

All possible numbers are formed using the digit
1,1,2,2,2,2,3,4,4 taken all at a time. The number
of such numbers in which the odd digits occupy
even place is

Answer 1

$\large\underline\purple{\bold{Solution :- }}$

We know that

$\boxed{\sf \:^{n}P_r=\dfrac{n!}{(n-r)!}}$

☆ From the question, it is given that all the digits 1, 1, 2, 2, 2, 2, 3, 4, 4 are used and all the digits are taken at a time.

☆ So, we should form 9 digit number.

☆ In these 9 digits, we have 6 even digits and 3 odd digits.

☆ It is also given that odd digit must occupy even place.

☆ In 9 digit number, we have 5 odd places and 4 even places. So, we should place 3 odd digits (1, 1, 3) in 4 even places.

☆ The number of ways of such arrangement in which 1's repeat twice is given by

$\tt \:  number \: of \: ways \: = \dfrac{\:^{4} P_3}{2!} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \tt \:  \dfrac{4!}{2! \times (4 - 3)!} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \dfrac{4 \times 3 \times 2!}{2! \times 1} \\ \: \: \: \: \: \: \: \: \ = 12$

☆ Now, we should place the remaining digits 2, 2, 2, 2, 4, 4 in 6 remaining places. In these, 2's repeat 4 times and 4's repeat twice, then

$\tt \: Number\:of\:ways = \dfrac{\:^{6} P_6}{2! \times 4!}$

$\tt \:  = \dfrac{6!}{2! \times 4!}$

$\tt \:  = \dfrac{6 \times 5 \times 4!}{4! \times2 \times 1 }$

$\tt \:  = 15$

☆ So, total number of ways in which the odd digits occupy

even place is 15 × 12 = 180 ways.

$\large \boxed{ \pink{ \bf \: Hence, number\:of\:ways = 180}}$