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junior college 2 | H2 Maths
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Date Posted: 7 months ago
Views: 145
1) Equivalent to arranging 1P 3A 1Y: 5!/3!
2i) 7C3 × 6C3
2ii) Using the complement method:
Total ways with no restrictions - both A and B are chosen = 7C3 × 6C3 - 6C2 × 5C2
3i) Equivalent to selecting 4 letters from 7: 7P4
3ii) Equivalent to selecting 1 of 2 pairs of repeated letters and 2 of 5 non-repeating letters: 2C1 × 5C2 × 4!/2!
3iii) 3 possible cases, no repeats, 1 repeat and 2 repeats: 7P4 + 2C1 × 5C2 × 4!/2! + 4!/(2!2!)
2i) 7C3 × 6C3
2ii) Using the complement method:
Total ways with no restrictions - both A and B are chosen = 7C3 × 6C3 - 6C2 × 5C2
3i) Equivalent to selecting 4 letters from 7: 7P4
3ii) Equivalent to selecting 1 of 2 pairs of repeated letters and 2 of 5 non-repeating letters: 2C1 × 5C2 × 4!/2!
3iii) 3 possible cases, no repeats, 1 repeat and 2 repeats: 7P4 + 2C1 × 5C2 × 4!/2! + 4!/(2!2!)
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Date Posted:
7 months ago
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