The number of permutations, of the digits 1, 2, 3, ..., 7 without repetition, which neither contain the string 153 nor the string 2467, is \_\_\_\_\_\_\_\_\_\_\_.
Answer: 4898
Solutions · 1
Official solution
Given that digits are 1,2,3,4,5,6,7
Total permutations =7!
Let p= Number which containing string 153
q= Number which containing string 2467
∴n(p)=5!×1⇒n(q)=4!×1⇒n(p∩q)=2!∴n(p∪q)=n(p)+n(q)−n(p∩q)=5!+4!−2!=120+24−2=142∴n (neither string 143 nor string 2467)
=7!−142=5040−142=4898