34
In how many different ways can the numbers ?256974? be arranged, using each digit only once in each arrangement, such that the digits 6 and 5 are at the extreme ends in each arrangement ?
- A48
- B720
- C36
- D360
Answer not set
Discussion 1
Log in or create a free account to reply and vote.
0
CCommunity
Solution
Case 1 : 6 at left end and 5 is at right end : 6 _ _ _ _ 5
Now, four empty places can be filled by 2,9,7 and 4 in = $4!$ ways
= $4 \times 3 \times 2 \times 1 = 24$
Case 2 : 6 at right end and 5 at left end : 5 _ _ _ _ 6
Similarly, no. of ways = $4!$
= $4 \times 3 \times 2 \times 1 = 24$
$\therefore$ Total no. of ways = $24 + 24 = 48$
Case 1 : 6 at left end and 5 is at right end : 6 _ _ _ _ 5
Now, four empty places can be filled by 2,9,7 and 4 in = $4!$ ways
= $4 \times 3 \times 2 \times 1 = 24$
Case 2 : 6 at right end and 5 at left end : 5 _ _ _ _ 6
Similarly, no. of ways = $4!$
= $4 \times 3 \times 2 \times 1 = 24$
$\therefore$ Total no. of ways = $24 + 24 = 48$