Dear all,
I posted this on mgmat forum to no avail.
I have a question about a question on Combinatroics Strategy in the textbook.
On pg. 68, there is a sample question:
An office manager must choose a five-digit lock code for the office door. The first and last digits of the code must be odd, and no repetition of digits is allowed. How many different lock codes are possible?
The explanation states that I should think of solving the question this way:
a - first digit, only odd
e - last digit, only odd
a b c d e
5 4
5 8 7 6 4
so, the total number of lock codes is therefore 5x8x7x6x4=6720
My question, by automatically assigning the e as 4, aren't we falsely assuming that d will not be an odd number? for example if a = 1, then e can only be e=3, 5, 7, 9. But how can we be so certain that d doesn't contain 3, 5, 7, 9? the only thing I can think of is that the five digit selection process begins with a, then e, then b, c, and d. That way, e can be 4. otherwise, e can't be 4.
Please help me clarify this.
I posted this on mgmat forum to no avail.
I have a question about a question on Combinatroics Strategy in the textbook.
On pg. 68, there is a sample question:
An office manager must choose a five-digit lock code for the office door. The first and last digits of the code must be odd, and no repetition of digits is allowed. How many different lock codes are possible?
The explanation states that I should think of solving the question this way:
a - first digit, only odd
e - last digit, only odd
a b c d e
5 4
5 8 7 6 4
so, the total number of lock codes is therefore 5x8x7x6x4=6720
My question, by automatically assigning the e as 4, aren't we falsely assuming that d will not be an odd number? for example if a = 1, then e can only be e=3, 5, 7, 9. But how can we be so certain that d doesn't contain 3, 5, 7, 9? the only thing I can think of is that the five digit selection process begins with a, then e, then b, c, and d. That way, e can be 4. otherwise, e can't be 4.
Please help me clarify this.
