P/C para

1)

here we have 2 I`s and 2 F`s and 3 E`s and 1 C and

1 T,1 N,1 V.

so we have (I,F,E,C,T,N,V).

four letters can be picked from these in 7C₄ ways and these can be arranged in thoose 4 places in 4! ways.

so total no : such words in which all the four letters are different is 7C₄.4!

remaning 2 i am really getting very long (but easy) arguments.... i am really not sure if thats the way to proceed.

like for 2)

removing I and C we have 2 F and 3E,1T.

here we can take 2 cases a)T included and b)T not included.

and again in both we will get subcases... but as the other numbers are less, i guess we can gamble out something....... continuing upon like this....

first one ive done .. but
the options were
Q2. (a) 1680 (b) 7921 (c)126 (d)144
Q3. (a) 504 (b)630 (C)1680 (d) none of these

bhargav.
in Q1. why have u ignored the letters N and V??

i thot it wud be ⁷C₄.4!

ya sorry,, i didnt notice them u r right..

ok for 3rd

2 vowels ca be in the following ways

a)2 I

b)1I , 1E

c)2E

case 1) 2 I

now 2 I`s can be picked in 1 way.

now for the next 2 consonants, if we pick up 2F then this can also be done in one way. and the whole 4 can be arranged in 4!/2!.2! ways

if we leave F, then we have 4 consonants to select from which can be done in 4C₂ ways.
and the whole 4 letters can be arranged in 4!/2! ways.
so total no:iof such words is 4C₂.4!/2!

now same is the case with 2 E`s

b) 1 E, 1 I

1 E can be picked in 1 way and 1 I can be picked in 1 way.

now for consonatns if 2 F`s are picked then we hav e total of 4!/2! words

but if F are not taken into consideration we have
remaining consonants can be selected in 4C₂ ways. and all these can be arranged in 4! ways.
so total 4C₂.4! such words.

now add this all up to get the answer.

seems long explanation but very easy to do and should take less than 2 min...

it means if the greatest coefficient is say 34 , then diff of its digits is 4-3=1 or -1 the other way...

if its 3 digit then i don think that makes anysense