Get Started With Combinatorics 4 – More Parade

In general, for $n$ people, there are $1 \times 2 \times 3 \times \ldots \times (n-1) \times n$ ways to arrange them.

From the previous question, when we have 7 people, there are 5040 ways to arrange them.
When we have 8 people, there are $8 \times 5040 = 40320$ ways to arrange them.
When we have 9 people, there are $9 \times 40320 = 362880$ ways to arrange them.
When we have 10 people, there are $10 \times 362880 = 3628800$ ways to arrange them.
When we have 11 people, there are $11\times 362880 = 39916800$ ways to arrange them.
When we have 12 people, there are $12 \times 39916800 = 479001600$ ways to arrange them.
When we have 13 people, there are $13 \times 479001600 = 6227020800$ ways to arrange them.

Hence, the minimum number of participants to reach more than a billion possible orderings, is 13.

I looked up factorial list in wikipedia. 13 was first one to exceed a billion

If we calculate we can see that 13! is the first factorial to give a value more than a billion.

sorry guys - you HAVE ta calculate this one. 1 billion means 9 zeroes. So , starting from....like 8 or 9 work your way upwards or work your way downwards from...maybe 99(glad i had a calculator, 99 was the first one that came into my mind, i tried 15 next, funnily enough) till you get to 13. Its approximately 6 times a billion. But its the 1st one to have 9 zeroes, so there's your answer. Almost zero brainwork involved. Cheers.

the arrangement of 'n' number of DISTINCT objects can be done in n! ways.(n! denotes the factorial of the integer) if we need arrangements more than one million, the factorial(n x (n-1) x (n-2) x .. . . . . . . x 1) should be greater than or equal to one million. thus thirteen is the first integer whose factorial exeeds one million.

13! is the first number whose factorial is over billion. (1,000,000)

For 12 people, 12 ! = 479001600
for 13 people, 13 ! = 6227020800 which is more than 1 billion...

Hence 13 is minimum possible people in parade