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Four letter words are formed by using the letters of the word INEFFECTIVE
Q1. The number of such words in which all the four letters are different is _____
Q2. The number of such four lettered words in which neither C nor I is present is _______
Q3. number of such four lettered words which contain two vowels and two consonants is _____
391)
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!
39remaning 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....
106first 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
106bhargav.
in Q1. why have u ignored the letters N and V??
i thot it wud be 7C4.4!
11Can the second one be approached like this
No. of required cases=Total no. of cases- (cases with 2I's 1C)-(cases with 1I and 1C)- (cases with 1I no C) - (cases with 1C and no I)
39ok 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...
106Q2. If the sum of coefficients in the expansion of (x-2y+3z)n is 128, then the difference of the integers in the greatest coefficient in the expansion of (1+x)n is _____
what do the bolded words mean?
Q3. AB is a focal chord of x2-2x+y-2=0 whose focus is S. If AS= 5/19, then SB is equal to _____
39it 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
393) in y2=4ax
bring the equation to this form.. then u have SA , a , SB are in harmonic progression
