Four digit numbers with all digits distinct are formed using the digits $1,2,3,4,5,6,7$ in all possible ways…

Four digit numbers with all digits distinct are formed using the digits $1,2,3,4,5,6,7$ in all possible ways. If $p$ is the total number of numbers thus formed and $q$ is the number of numbers greater than 3400 among them, then $p: q=$
  1. $3: 2$
  2. $4: 3$
  3. $6: 5$
  4. $7: 4$

Solution

$p=$ Total four digit number's formed using $1,2,3,4$, $5,6,7={ }^7 C_4 \times 4!=\frac{7!4!}{3!4!}=840$ For number greater than 3400 , Numbers starting with 3,4 are $={ }^5 C_2 \times 2!=20$ Now, first digit can be $4,5,6,7={ }^4 C_1 \times{ }^6 C_3 \times 3!=480$ Numbers starting with 3 and second digit 5, 6, 7 are $={ }^3 C_1 \times{ }^5 C_2 \times 2!=\frac{3!\times 5!}{3!2!}=60$ $\therefore q=$ Total numbers greater than $3400=20+480+60=560$ Now, $p: q=840: 560=3: 2$.

Asked in: AP EAMCET 2024 (21 May Shift 2)

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