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


1. $2240$
2. $2296$
3. $2620$
4. $4536$
* 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: $B$
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4g bands 
4g fdd-lte: b1(2100),b2(1900),b3(1800),b4(aws),b5(850),b7(2600),b8(900),b12(700),b13(700), b18(800),b19(800),b20(800),b25(1900),b26(850),b28(700) / 
4g tdd-lte: b34(2010),b38(2600),b39(1900),b40(2300),b41(2500... tallies are usually marked in a bunch of 4. finding the correct answer here is important, but perhaps equally important is finding it using the method of matrix reduction, since those are the instructions.the augmented matrix is4 -2 -5 -514 -1 -2 -364 -2 4 -24you'll have to forgive the poor formatting. this answer box doesn't have a proper math symbol editor.using elementary row operations, the reduced row echelon form of the augmented matrix is1 0 0 -60 1 0 60 0 1 3from which we conclude that the system has a unique solution (-6,6,3). answer: `1/221`
solution: n/a answer: `4+1+(1)/(4)+(1)/(6)+...`
solution: n/a we know that dividend = divisor × quotient + remainderaccording to the problem :dividend 1 = 245dividend 2 = 1029dividend - remainder = divisor × quotientso dividend 1- remainder = 240 = divisor × quotient 1prime factor of 240 = 24 × 3 × 5dividend 2 – remainder = 1024 = divisor × quotient 2prime factor of 1024 = 24 × 26since, the divisor is common for both the numbers we need to find the highest common factor between both the numbers. from the prime factors, we find thehighest common factor between the two numbers is 24 = 16 of three numbers, second is twice the first and is also thrice the third. if the average of the three numbers is 44, the largest number is:

option 1) 24                              option 2) 56option 3) 72               option 4) 108option 5)    36 2, 0 ( any two integers with one of them as 0). i = ke/r
30 = 120k/16
k=4
now you can answer the question when you have fixed the typos. a - ii, b - iv, c - i, d - iii