In how many ways can the digits 2,3,5,7 and 9 be placed to form a three-digit number so that the higher order digit is always greater than the lower order digits? (assume digits are all different
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It is equal to the selection of 3 objects from 5 distinct objects which is equal to
5C3 =10 .
Because if we select 3 digits from 5 digits so for each combination of selection three digits there will be only one way in which higher order digit is followed by lower order digits .
for eg if we select 9,7,5 then we can arrange it 6 different ways
975 ,579,759,957,795,597 .Out of these arrangements we will take 975 .so it is equal to selection of 3 digits from 5 digits .
5C3 =10 .
Because if we select 3 digits from 5 digits so for each combination of selection three digits there will be only one way in which higher order digit is followed by lower order digits .
for eg if we select 9,7,5 then we can arrange it 6 different ways
975 ,579,759,957,795,597 .Out of these arrangements we will take 975 .so it is equal to selection of 3 digits from 5 digits .
Answered by
8
Answer:
Step-by-step explanation:
We are having 5 digits : 2,3,5,7,9
These are to be arranged in such a way thay digit at tens place is to be greater than ones place and digit at hundredth place is greater than both ones and tens.
So now
Taking 2 at hundredth position : we can't make any 3digit number which will satisfy above conditions
Taking 3 at hundredth : it will also not contribute in making a 3 digit number
Taking 5: 532 is the only 1 number possible
Taking 7: 753,752,732 i.e total 3 numbers are possible.
Taking 9: 975,973,972,953,952,932 i.e. total 6 numbers are possible.
Therefore total numbers possible are = 1+ 3 + 6=10
Answer= 10
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