Q87 / 180 · KCET 2019 · Mathematics
The number of \( 4 \) digit numbers without repetition that can be formed using the digits \( 1,2,3,4 \),
\( 5,6,7 \) in which each number has two odd digits and two even digits is
\( 5,6,7 \) in which each number has two odd digits and two even digits is
- A\( 436 \)
- B\( 454 \)
- Correct answerC\( 432 \)
- D\( 450 \)
Solution
(C)
Given digits are \( 1,2,3,4,5,6,7 \).
Two even digits can be selected in \( { }^{3} C_{2} \)
Two odd digits can be selected in \( { }^{4} C_{2} \) ways.
These selected \( 4 \) digits can be arranged in \( 4 ! \) ways.
\( \therefore \) Total number of ways \( ={ }^{4} C_{2}{ }^{3} C_{2}{ }_{2} ! \)
\( =6 \times 3 \times 24 \)
\( =18 \times 24 \)
\( =432 \)
Given digits are \( 1,2,3,4,5,6,7 \).
Two even digits can be selected in \( { }^{3} C_{2} \)
Two odd digits can be selected in \( { }^{4} C_{2} \) ways.
These selected \( 4 \) digits can be arranged in \( 4 ! \) ways.
\( \therefore \) Total number of ways \( ={ }^{4} C_{2}{ }^{3} C_{2}{ }_{2} ! \)
\( =6 \times 3 \times 24 \)
\( =18 \times 24 \)
\( =432 \)
Hence, the correct answer is option: 'C'.