The number of 6 digit numbers that can be formed from 0 1 5 6 7 and 8 in which the first digit is not 0 are. how many numbers are there?
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Let the six digit number be termed as ABCDEF
AA cant take 00 (if it takes 00 then the number will be five digit) but can choose all other 1,5,7,1,6,5,7,So 55 ways.
BB cant choose AA since any number shouldn't be repeated, so among this 55 numbers one will not b there in the choice of BB, but BB can choose 0 also. Hence 55 ways.
CC can't choose AA and BB. Ie, except this two but all the other 44. So 44 ways
DD cant choose AA, BB abd CC. 33 choices lost. Rest 3 DD can choose. So 33 ways
EE cant choose AA, BB, CC and DD. So left 22 choices. So 22 ways
FF cant choose all AA, BB, CC, DD and EE Left only one choice. So 11 way
So in total there are 5∗5∗4∗3∗2∗1=6005∗5∗4∗3∗2∗1=600six digit numbers formed by {0,1,3,5,7,90,1,3,5,7,9} without repetition.
Among this 10 divides only those which ends with 00. So we are fixing 00 at FF.
So we are left with only five numbers, and it could be any of {1,5,7,91,3,5,7,9} because divisibility by 10 needs only end digit 00
So as we proceed above there will be 5∗4∗3∗2∗1∗1=1205∗4∗3∗2∗1∗1=120 numbers where each number in the product is the number of choices for A,B,C,D,E,FA,B,C,D,E,F respectively.
Therefore 600600 numbers of which 120120 are divisible by 100
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AA cant take 00 (if it takes 00 then the number will be five digit) but can choose all other 1,5,7,1,6,5,7,So 55 ways.
BB cant choose AA since any number shouldn't be repeated, so among this 55 numbers one will not b there in the choice of BB, but BB can choose 0 also. Hence 55 ways.
CC can't choose AA and BB. Ie, except this two but all the other 44. So 44 ways
DD cant choose AA, BB abd CC. 33 choices lost. Rest 3 DD can choose. So 33 ways
EE cant choose AA, BB, CC and DD. So left 22 choices. So 22 ways
FF cant choose all AA, BB, CC, DD and EE Left only one choice. So 11 way
So in total there are 5∗5∗4∗3∗2∗1=6005∗5∗4∗3∗2∗1=600six digit numbers formed by {0,1,3,5,7,90,1,3,5,7,9} without repetition.
Among this 10 divides only those which ends with 00. So we are fixing 00 at FF.
So we are left with only five numbers, and it could be any of {1,5,7,91,3,5,7,9} because divisibility by 10 needs only end digit 00
So as we proceed above there will be 5∗4∗3∗2∗1∗1=1205∗4∗3∗2∗1∗1=120 numbers where each number in the product is the number of choices for A,B,C,D,E,FA,B,C,D,E,F respectively.
Therefore 600600 numbers of which 120120 are divisible by 100
please mark as brainlist
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