sana.noor wrote:How many 7-digit even numbers less than 3,000,000 can be formed using the following digits: 1, 2, 2, 3, 5, 5, 6?
a) 30
b) 60
c) 210
d) 180
e) 300
OA is C
When a permutation includes IDENTICAL elements, we must DIVIDE by the number of ways the identical elements can be arranged.
Example:
Number of ways to arrange AABBB = 5!/(2!3!).
Here, we divide by 2! to account for the 2 identical A's and by 3! to account for the 3 identical B's.
In the problem above, the 1st digit must be 1 or 2 (since the integer must be less than 3,000,000) and the last digit must be 2 or 6 (since the integer must be even).
Case 1: 1st digit = 2, last digit = 2
Number of ways to arrange the remaining five digits 1, 3, 5, 5, 6 = 5!/2! = 60.
Case 2: 1st digit = 2, last digit = 6
Number of ways to arrange the remaining five digits 1, 2, 3, 5, 5 = 5!/2! = 60.
Case 3: 1st digit = 1, last digit = 2
Number of ways to arrange the remaining five digits 2, 3, 5, 5, 6 = 5!/2! = 60.
Case 4: 1st digit = 1, last digit = 6
Number of ways to arrange the remaining five digits 2, 2, 3, 5, 5 = 5!/(2!2!) = 30.
Total ways = 60+60+60+30 = 210.
The correct answer is
C.
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