How many numbers can be formed from 1,2 3 4 5 without repetition when the digit at the units place be greater than that in the tens place 54?

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when we take 1 as tenth place digit then in units place 2,3,4,5 can be used satisfying the condition. so leftmost three digits(as it is non repetitive)can be combined into 3! way total ways=4*3! for 2=>>3*3! for 3=>>2*3! for4=>>1*3! therefore total numbers will be=(4+3+2+1)*3!=60
option (c) is correct
11 years agoHelpfull: Yes(12) No(1)
c) 60 taking separate cases for 1 at tenth place 4 ways of filling unit place ,total=4*3*2*1=24 for 2 at tenth place 3 ways of filling unit place , total=3*3*2*1=18 for 3 at tenth place 2 ways of filling unit place , total=2*3*2*1=12 for 4 at tenth place 1 ways of filling unit place , total=1*3*2*1=6 for 1 at tenth place 0 ways of filling unit place
TOTAL=60
11 years agoHelpfull: Yes(7) No(0)
c)60 The nos. have to be 5 digit nos., since we are using all the five digits. There are 10 ways to arrange nos. in ten's and unit's place to satisfy the condition in question ... 21,31,41,51,32,42,52,43,53,54 So now we have 10 possible combos for the last two digits: _ _ _ X X We are left with the first 3. There are 3 nos. left to fill these positions since 2 are getting used to fill up the last two positions. These 3 nos. can be arranged in these positions in 3P3 ways = 6 ways.
So final no. of combinations = 10*6 = 60
11 years agoHelpfull: Yes(6) No(1)
For the condition to be satisfied, we need 2 nos out of 5, which can be chosen as 5C2 ways (gives 10), and the remaining three can be arranged between themselves which is 3! ways( gives 6). Now multiply both 5C2 * 3! = 60.
Hope it helps..
7 years agoHelpfull: Yes(3) No(0)
60 ...1. --> 1*2*3*1*4 = 6*4 ...2. --> 1*2*3*1*3 = 6*3 ...3. --> 1*2*3*1*2 = 6*2 ...4. --> 1*2*3*1*1 = 6*1 ...5. --> 1*2*3*1*0 = 6*0
Total = 6*(0+1+2+3+4) = 60
11 years agoHelpfull: Yes(1) No(2)
60 is answer
4 years agoHelpfull: Yes(1) No(3)