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8-digit numbers are formed using the digits 1, 1, 2, $2,2,3,4,4$. The number of such numbers in which the odd digits do no occupy odd places, is:
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The correct answer is:
120
120
In 8 digits numbers, 4 places are odd places.
Also, in the given 8 digits, there are three odd digits 1,1 and 3.
No. of ways three odd digits arranged at
$$
\text { four even places }=\frac{4 P_3}{2 !}=\frac{4 !}{2 !}
$$
No. of ways the remaining five digits 2 , 2, 2, 4 and 4 arranged at remaining five places $=\frac{5 !}{3 ! 2 !}$
Hence, required number of 8 digits number $=\frac{4 !}{2 !} \times \frac{5 !}{3 ! 2 !}=120$
Also, in the given 8 digits, there are three odd digits 1,1 and 3.
No. of ways three odd digits arranged at
$$
\text { four even places }=\frac{4 P_3}{2 !}=\frac{4 !}{2 !}
$$
No. of ways the remaining five digits 2 , 2, 2, 4 and 4 arranged at remaining five places $=\frac{5 !}{3 ! 2 !}$
Hence, required number of 8 digits number $=\frac{4 !}{2 !} \times \frac{5 !}{3 ! 2 !}=120$
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