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Number of permutations of 1,2,3,4,5,6,7,8, and 9 taken all at a time are such that digit 1 appearing somewhere to the left of 2 and digit 3 appearing to the left of 4 and digit 5 somewhere to the left of 6 , is

 (e.g.815723946 would be one such permutation)

 

Option: 1

9.7 !


Option: 2

8 !


Option: 3

5 ! .4 !


Option: 4

8 ! .4 !


Answers (1)

Number of digits are 9 

Select 2 places for the digit 1 and 2 in { }^9 C_2 ways

from the remaining 7 places, select any two places for 3 and 4 in { }^7 C_2 ways and from the remaining 5 places select any two for 5 and 6 in { }^5 C_2 ways.

Now, the remaining 3 digits can be filled in 3! ways.

\therefore Total ways ={ }^9 \mathrm{C}_2 \cdot{ }^7 \mathrm{C}_2 \cdot{ }^5 \mathrm{C}_2 \cdot 3 !

                      \begin{aligned} & =\frac{9 !}{2 ! \cdot 7 !} \cdot \frac{7 !}{2 ! \cdot 5 !} \cdot \frac{5 !}{2 ! \cdot 3 !} \cdot 3 ! \\ \\& =\frac{9 !}{8}=\frac{9 \cdot 8 \cdot 7 !}{8}=9 \cdot 7 ! \end{aligned}

Posted by

Ramraj Saini

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