JEE Main 2020 · 7 January · morning · Single correct · +4 / −1
Total number of 6-digit numbers in which only and all the five digits 1, 3, 5, 7 and 9 appear, is :
- (a) 5/2( 6! )
- (b) 6!
- (c) 5⁶
- (d) 1/2( 6! )
Answer and solution
Answer (a) 5/2( 6! )
Solution
Maths sir · 3 steps
Six places but only five digits, so exactly one digit must appear twice. Pick it first, then arrange.
- Which digit repeats. All five digits must appear, which fills 5 of the 6 places. The sixth place must reuse one of the five digits, so exactly one digit appears twice. That digit can be chosen in 5 ways.
- Arrange the six. Say 1 is repeated, so the digits are 1, 1, 3, 5, 7, 9. Arranging 6 objects with one pair identical gives 6!/2! arrangements, because swapping the two 1s gives the same number.
Total = 5 × 6!/2! = 5/2(6!) = 1800
- So. None of the digits is 0, so every arrangement is a valid 6-digit number. The count is 5/2(6!), which is option (a).
The idea to keep. When n places must use all of k digits and n is just above k, first decide which digits repeat, then arrange with the identical-objects formula n!/(p! q!…).
Answer from the NTA final answer key. Solution by PrepWiser.