How many $4$-digit positive integers exist that satisfy the following conditions: (A) Each of the first two digits must be $1$,
$4$, or $5$, and (B) the last two digits cannot be the same digit, and (C) each of the last two digits must be $5$, $7$, or $8$
1 answer:
Answer:
54 possibles
Step-by-step explanation:
Digit ONE 3 choices
Digit two 3 choices
digit three 3 choices
digit four 2 choices
3 x 3 x 3 x 2 = 54 choices meet all of the conditions
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