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Worked Examples · Example 4

Q.Verify that 7P3=7C3×3!^{7}P_{3} = {}^{7}C_{3} \times 3!.

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Left side — compute 7P3^{7}P_{3} directly:

7P3=7×6×5=210^{7}P_{3} = 7\times6\times5 = 210

Right side — compute 7C3^{7}C_{3} first, then multiply by 3!3!:

7C3=7!3! 4!=7×6×5×4!3!×4!=7×6×53!=2106=35^{7}C_{3} = \frac{7!}{3!\,4!} = \frac{7\times6\times5\times4!}{3!\times4!} = \frac{7\times6\times5}{3!} = \frac{210}{6} = 35

7C3×3!=35×6=210^{7}C_{3}\times3! = 35\times6 = 210 …

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