jaya vignesh according to my understaning i will try to explain MR. J correct me if i am wrong

so there are _ _ _ _ _….5 places you can’t enter 0 in the 1st position if u did so it will become 4 digit number. so possible number are (1,2,3,4) in first place
from second place on words wee can enter 0 and any three number from (1,2,3,4) since one number is placed in first place.so 4! ways from second place on words.
so 4*4!=96 ways.
in first place each letter can occur 24 time and from second letter onward only 18 times because of 0. then as usual

can we solve the last question by subtracting addition of all 4 digit numbers formed by 1,2,3,4 from addition of all five digit numbers formed by 0,1,2,3,4,5?? so this will actually give all the five digit numbers.

Try to think for yourself rather than just asking for solutions! Else the whole point of this blog is defeated. Here, if 1 is in the first place, the other 4 digits need to be arranged in the other 4 places. Ask yourself in how many ways this can be done…

You should be able to reason it out from the three examples given. So I am not going to provide a solution (or answer) for that! If it doesn’t force you to think for yourself, this blog will be no different from any board textbook…

Will it be solved like this:
(625+500+500+500+500+500)(111111)(1+2+3+4)
because 1st place each digit will occur (2500/4) times,
and and in further places, each will come 2500/5 times as repetition is allowed?

sir, I ve in this different way. in the last qn there are 96 nos. out of which each of 1,2,3,4 appears in 10000 place for 24 times. so in rest of 72 nos starting with 1,2,3 it ll appear in each of the places for 72/4 times which is 18.

In the first question , if instead of 1,2,3,4,5 …if there are 4 numbers and we have to form 5 digit numbers .. then how can i proceed ??

Dear J,

I didn’t understand the last question :(.

Could not understand the last question.We require a better explanation

i doubt what radhika asked can’t be done is that possible ?

jaya vignesh according to my understaning i will try to explain MR. J correct me if i am wrong

so there are _ _ _ _ _….5 places you can’t enter 0 in the 1st position if u did so it will become 4 digit number. so possible number are (1,2,3,4) in first place

from second place on words wee can enter 0 and any three number from (1,2,3,4) since one number is placed in first place.so 4! ways from second place on words.

so 4*4!=96 ways.

in first place each letter can occur 24 time and from second letter onward only 18 times because of 0. then as usual

can we solve the last question by subtracting addition of all 4 digit numbers formed by 1,2,3,4 from addition of all five digit numbers formed by 0,1,2,3,4,5?? so this will actually give all the five digit numbers.

Yes, that is a valid approach ðŸ™‚

regards

J

Sir,pls explain the last question in detail.

1.How the first digit occur 24 times

Try to think for yourself rather than just asking for solutions! Else the whole point of this blog is defeated. Here, if 1 is in the first place, the other 4 digits need to be arranged in the other 4 places. Ask yourself in how many ways this can be done…

regards

J

what if in last question repetition is allowed?

You should be able to reason it out from the three examples given. So I am not going to provide a solution (or answer) for that! If it doesn’t force you to think for yourself, this blog will be no different from any board textbook…

regards

J

Will it be solved like this:

(625+500+500+500+500+500)(111111)(1+2+3+4)

because 1st place each digit will occur (2500/4) times,

and and in further places, each will come 2500/5 times as repetition is allowed?

I already said, I am not going to give an answer!

regards

J

at least a yes or no ðŸ˜›

sir, I ve in this different way. in the last qn there are 96 nos. out of which each of 1,2,3,4 appears in 10000 place for 24 times. so in rest of 72 nos starting with 1,2,3 it ll appear in each of the places for 72/4 times which is 18.