@rkshtsurana said:me getting 5
7C4 - 5C2 nhi hoga kya ? :(
@soumitrabengeri said:even i am getting the answer as 5..seems wrong..but can't figure out where
bhai un dono ko hata do..ab 5 me 4 lelo ==5c4 = 5
@rkshtsurana said:bhai un dono ko hata do..ab 5 me 4 lelo ==5c4 = 5
correct..same method used by me
@surajmenonv said:wats the relevance of P(n) = n+3???? and @krum : y did u assume quadratic? its only said polynomial
it was given p(n) has 2 distinct solutions so i assumed it to be quad.
@surajmenonv said:chill man.....abhi to season hardly shuru hua h....the thread will pick up speed gradually...
ham toh chill he rahte hein.. yar..
@krum said:p(x)=ax^2+bx+c=>10=289a+17b+c=>17=576a+24b+c==>287a+7b=7==>41a+b=1==>(a,b)= (-1,42)/(1,-40)p(n)=an^2+bn+c==>an^2+bn+c=n+3==>an^2+(b-1)n+c-3=0sum of roots = (1-b)/a = (1-42)/-1 or (1+40)/1 = 41
hmm sahi hai bhai..mast bana rhe ho..
@soumitrabengeri said:@sujamait Ye lo bhai..IIFT 2006 questionIf a is not equal to n*(pi) and tan(a) is the GM of sina and cosa, then find the square of..2-4sin^2(a)+3sin^4(a)-sin^6(a)1) 12) 43) 1/44) NOT
tan^2a=sina . cosa
sina = cos^3a
kuch bat nhn bani..
@rachit_28 said:Ben and Ann are among 7 contestants from which 4 semifinalists are to be selected. Of the different possible selections, how many contain neither Ben nor Ann?I am getting 25. Please confirm if it's right.
5C4 = 5
@raakesh88 said:H.C.F. of three positive integers €“ A, B and C €“ is 6. If the sum of the three integers is 48, how many triplets (A, B and C) are possible?
6( a+b+c)=48
a+b+c =8
now need to select all 3 as co prime
1,1,6
1,2,5
1,3,4
2,2,4 not possible so remove it
2,3,3
so 4 triplets , if order doesnt matter else
3+6+6+3 =18
@rkshtsurana said:directing to @krum
??
@rachit_28 said:7C4 - 5C2 nhi hoga kya ?
u are missing cases where only B or only A were included
=>7c4-5c2-5c3-5c3=35-10-10-10=5
=>7c4-5c2-5c3-5c3=35-10-10-10=5
@soumitrabengeri said:@sujamait Ye lo bhai..IIFT 2006 questionIf a is not equal to n*(pi) and tan(a) is the GM of sina and cosa, then find the square of..2-4sin^2(a)+3sin^4(a)-sin^6(a)1) 12) 43) 1/44) NOT
is it 1?
@techsurge said:6( a+b+c)=48a+b+c =8now need to select all 3 as co prime1,1,61,2,51,3,42,2,4 not possible so remove it2,3,3so 4 triplets , if order doesnt matter else3+6+6+3 =18
han yaar, i ignored 2,2,4 - 3 cases 
@rachit_28 said:Ben and Ann are among 7 contestants from which 4 semifinalists are to be selected. Of the different possible selections, how many contain neither Ben nor Ann?I am getting 25. Please confirm if it's right.
shud be 5
@rachit_28 said:Ben and Ann are among 7 contestants from which 4 semifinalists are to be selected. Of the different possible selections, how many contain neither Ben nor Ann?I am getting 25. Please confirm if it's right.
all pairs =7c4
containing ben 6c3
containg ann 6c3
cotaining both 5c2
my soln = 7c4 -6c3-6c3+5c2
35-20-20+10=5
@soumitrabengeri said:@sujamait Ye lo bhai..IIFT 2006 questionIf a is not equal to n*(pi) and tan(a) is the GM of sina and cosa, then find the square of..2-4sin^2(a)+3sin^4(a)-sin^6(a)1) 12) 43) 1/44) NOT
1
(tana)2=sina * cosa
=>sina=cosa3
sina2=(cosa2)3
sina2=(1-sina2)3
solving this we get,
3sina4-4sina2-sina6=-1
so 2-4sin^2(a)+3sin^4(a)-sin^6(a)=2-1=1
@raakesh88 said:H.C.F. of three positive integers €“ A, B and C €“ is 6. If the sum of the three integers is 48, how many triplets (A, B and C) are possible?
shud be 4. without arrangement.
@soumitrabengeri said:@sujamait@maddy2807 Here is the solution to the problem
i headed in opposite direction. do u hv more IIFT questions
@maddy2807 said:i headed in opposite direction. do u hv more IIFT questions
I do..will be posting some soon
@rachit_28 said:Ben and Ann are among 7 contestants from which 4 semifinalists are to be selected. Of the different possible selections, how many contain neither Ben nor Ann?I am getting 25. Please confirm if it's right.
ye to 5c4 = 5 hona chahiye bas :splat:
@rachit_28 said:7C4 - 5C2 nhi hoga kya ?
yaar wo tab hota jab question is accepting the case of any single persons absence as well. here, it's saying both of them shouldn't be there. it's neither nor either ;)