
Daniel E. answered 05/19/14
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Mathematics Made Easy with Dr. Dan
For the first one we have 50 choices for there first number. Since the first number is picked we only have 1 choice for the second since it has to be the same as the first. Since the third number can be anything we once again have 50 choices. This gives us 50∗1∗50=2500 combinations.
For the second we have three different scenarios either the first two match and the third is different, the second two match and the first is different, or the first and third match and the second is different. In the first case we once again have 50 choices for the first number and 1 for the second, however we only have 49 choices for the third number since it can not be the same as the first two. Giving us 50∗1∗49=2450 combinations in option one. Since the other two options are very similar, each option will give you 2450 combinations and you have three options. All we need to do is add 2450+2450+2450=7350 combinations.