Question

The number 916238457 is an example of a nine-digit number which contains each of the digits 1 to 9 exactly once. It also has the property that the digits 1 to 5 occur in their natural order while the digits 1 to 6 do not. The number of such numbers is
A. 2268
B. 2520
C. 2975
D. 1560

Answer 1

Hint: We first try to place the digits 1 to 6 in such a way that 1 to 5 is in order but 1 to 6 isn’t. We pick 6 spots out of 9 and find a solution. Then we place the rest three digits to form the multiplication and find the solution.

Complete step-by-step answer:
The given condition for the arrangements of the 9 digits is that the digits 1 to 5 occur in their natural order while the digits 1 to 6 do not.
We first take 9 spots to fill-up and then take 1 to 6 to complete first.
To place them we first choose 6 spots out of 9. This can be done in $ {}^{9}{{C}_{6}} $ ways.
Now in those 6 places, spots for 1 to 5 are fixed and we also need to be careful about the position of the digit 6 as it can’t be placed at the end. That will create the numbers 1 to 6 being in order.
For 6 we have 5 places.
We are left with 3 spots out of 9 and 3 digits to be filled with.
The number of choices is $ 3! $ .
Therefore, the final solution is $ {}^{9}{{C}_{6}}\times 5\times 3!=2520 $ . The correct option is B.
So, the correct answer is “OPTION B”.

Note: We have to be careful about the explanation of the given conditions. If we place 6 at the end then the digits 1 to 6 remain in order. Therefore, permutation only happens for 6.