Question

how many number lying between 100 and 1000 can be formed with the digits 0,1,2,3,4,5,6,7 if repetition of digits is not allowed?

• A. ${}^7{p}_3-{}^6{p}_3$
• B. ${}^{10}{p}_4-{}^6{p}_2$
• C. ${}^7{p}_3-{}^6{p}_2$
• D. ${}^8{p}_3-{}^7{p}_2$

Answer 1

The correct option is D ${}^8{p}_3-{}^7{p}_2$

Solution :-

Number between 100 and 1000 will be 3 digit number

3 digit number will be formed from the digits

0,1,2,3,4,5,6,7

But there include number starting with'O' like 012,023 etc.

Required number = Total 3 digits - 3 digit number which have

0 in the beginning

Total 3 digit numbers = ${=}^n{P}_r{=}^8{P}_3=\frac{8!}{(8-3)!}=\frac{8!}{5!}=8\times 7\times 6$

$=56\times 6=336$

Total 3 digit numbers starting with 0 = ${}^7{P}_2=\frac{7!}{5!}=7\times 6=42$

Hence required number is = 336-72 = 264

${=}^8{P}_3{-}^7{P}_2$