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How many 2-digit numbers can be formed from the set of digits of the number $72372$?
My try :
The set of digits of $72372={{2,3,7}}$
The number of 2-digit numbers =$3×3=9$ "with repetition"
And $3×2=6$ "without repetition"
Is that right ?
$\endgroup$ 5
. How many two-digit even numbers can you form using the digits 0, 1 , 2, 3, 4, 5, 6, 7, 8, 9? a. repetition is allowed b. no repetition ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Of the 10 digits, 5 are even. So in both cases there are 5 choices for the units digit of the 2-digit number. (a) If repetition is allowed, then there are 10-1 = 9 choices for the tens digit. (9 because you can not use zero as the "tens" digit). The total number of possible 2-digit even numbers is 9*5 = 45, if repetition is allowed. (b) If repetition is not allowed, then there are two cases. Case 1. The "units" digit is zero. Then you have 9 choices for the "tens" digits. Hence, 9 (nine) 2-digit numbers are possible, ending by "0". Case 2. The "units" digit is any of 4 even digits 2, 4, 6 or 8. Then you have 8 choices for the "tens" digit. (Only 8, because you can not use zero and just used "units" digit). Hence, 8*4 = 32 2-digit numbers are possible in this case. The total for case 1 + case 2 is 9 + 32 = 41. So, in this problem / (sub-problem) the answer is: 41 2-digit even numbers are possible, if repetition is not allowed. -----------------
Notice that the answers to (a) and (b) are CONSISTENT:
From 45 numbers of the set (a) you need exclude four numbers 22, 44, 66 and 88 to get 41 number of the set (b).