A) 120
B) 300
C) 420
D) 20
Correct Answer: C
Solution :
The units place can be filled in 4 ways as any one of 0, 2, 4 or 6 can be placed there. The remaining three places can be filled in with remaining 6 digits in \[^{6}{{P}_{3}}\] = 120 way. So, total number of ways = 4 × 120 = 480. But, this includes those numbers in which 0 is fixed in extreme left place. Numbers of such numbers \[=3\times {{\,}^{5}}{{P}_{2}}\,=3\times 5\times 4=60\]0 | × | × | × |
Fix | \[^{5}{{P}_{2}}\] ways | 3 ways (only 2, 4 or 6) |
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