edited by
12,472 views
38 votes
38 votes

How many $4$-digit even numbers have all $4$ digits distinct?

  1. $2240$
  2. $2296$
  3. $2620$
  4. $4536$
edited by

9 Answers

Best answer
65 votes
65 votes
  • If the number ends with a $0$ then there are $9$ choices for the first digit, $8$ for the second and $7$ for the third, which makes $1\times9\times8\times7= 504$ possibilities.

  • If the number is even ending with something else than $0$ then there are $4$ choices for the last digit, $8$ choices for the first digit (no $0$ nor the last digit), $8$ for the second digit and $7$ for the third digit, which makes $4\times8\times8\times7 =1792$

Together, this gives $2296$ numbers with $4$ distinct digits that are even. Note that this does not allow leading $0$, as you see to want it based from the question.

Correct Answer: $B$

edited by
9 votes
9 votes

$Method\ 1:$

$-$ $-$ $-$ $0$ $9.8.7.1$
$-$ $-$ $-$ $2$ $8.8.7.1$
$-$ $-$ $-$ $4$ $8.8.7.1$
$-$ $-$ $-$ $6$ $8.8.7.1$
$-$ $-$ $-$ $8$ $8.8.7.1$

$9.8.7.1+(8.8.7.1)\times 4=2296$


$Method\ 2:$

$1)\ Total\ \#\ of\ 4\ digit\ numbers\ and\ all\ distinct=9.9.8.7=4536$

$2)\ \#\ of\ 4\ digit\ odd\ numbers\ and\ all\ distinct=8.8.7.5=2240$

$Even=Total-Odd=4536-2240=2296$

8 votes
8 votes
Available digits are { 0,1 , 2  ... 9}

Even { 0 2 4 6 8}

Odd { 1 3 5 7 9 }

Pos 4 must be one of even number . But Pos 1 can be odd or even but not 0.

(case 1 )

POS 1: ODD { ONE OF 1 3 5 7 9}

POS 2:7C1

POS:3 8C1                                                 SO Total is : 5 X (7x8x5)=1400

POS 4: ONE OF EVEN NO i.e. 5C1

(CASE 2: POS1 IS EVEN INLUDING 0)

POS 1: EVEN { ONE OF 0,2,4,6,8}

POS 2:7C1

POS:3 8C1                                                 SO Total is : 5 X (7x8x4)=1120

POS 4: ONE OF EVEN NO EXCLUDING THE DIGIT OF POS1 i.e. 4C1

(Case 3: POS1 IS 0)

POS 1: 0

POS 2:7C1

POS:3 8C1                                                 SO Total is : 1 X (7x8x4)=224

POS 4: ONE OF EVEN NO EXCLUDING 0  i.e. 4C1

The ans is : case1 + case2 - case3 =(1400+1120-224)=2296
6 votes
6 votes
I have tried in this way -

No of 4 digit even numbers= Total no of 4 digit even numbers -  Number having 0 at the front

   Total no of 4 digit even numbers =last place is filled in 2,4,6,8,0 ie 5 ways. Next place in 9 ways, next in 8 ways and next in 7 ways (go from right)

Number having 0 at the front = last place is filled in 2,4,6,8 ie 4 ways, next place in 8 ways and next in 7 ways and first place is for 0.

Required number of numbers = (7*8*9*5)-(7*8*4)  = 2520-224 = 2296
Answer:

Related questions

37 votes
37 votes
6 answers
4
Kathleen asked Sep 14, 2014
17,886 views
Which of the following statements is false?An unambiguous grammar has same leftmost and rightmost derivationAn LL(1) parser is a top-down parserLALR is more powerful than...