how many four digit numbers of the form PQRS are there with P is less than Q is less than R is less than S ?
189, 199, 126, 303
ans is 126
ca an1 pls give an illustrative approach i couldn't unddderstand the approach given in the book???
@quant89 said:how many four digit numbers of the form PQRS are there with P is less than Q is less than R is less than S ?189, 199, 126, 303ans is 126ca an1 pls give an illustrative approach i couldn't unddderstand the approach given in the book???
Possible=9*9*8*7
Favourable=1/24
As in arrangement of 4 numbers only one will satisfy your condition.
So total= 9*9*8*7/24= 189
Mera 189 aa rha.
Favourable=1/24
As in arrangement of 4 numbers only one will satisfy your condition.
So total= 9*9*8*7/24= 189
Mera 189 aa rha.
@quant89 said:how many four digit numbers of the form PQRS are there with P is less than Q is less than R is less than S ?189, 199, 126, 303ans is 126ca an1 pls give an illustrative approach i couldn't unddderstand the approach given in the book???
It will form up series
(6.7 + 5.6.2 + 4.5.3 + 3.4.4 + 2.3.5 + 1.2.5) / 2 + 1 => 126
bit lengthy qstn but can be done quickly with practice
(6.7 + 5.6.2 + 4.5.3 + 3.4.4 + 2.3.5 + 1.2.5) / 2 + 1 => 126
bit lengthy qstn but can be done quickly with practice
@quant89 said:how many four digit numbers of the form pqrs are there with p189, 199, 126, 303ans is 126ca an1 pls give an illustrative approach i couldn't unddderstand the approach given in the book???
@quant89 said:bhai type toh poora kia tha lekin post nhi hua ruk dobara krti hu...how many four digit numbers of the form pqrs are there with p189, 199, 126, 303ans is 126ca an1 pls give an illustrative approach i couldn't unddderstand the approach given in the book???
plz delete the unnecessary posts
@Ashmukh said:How many subsets of {1, 2, 3 €Ś 11} contain at least one even integer?(1) 1900 (2) 1964(3) 1984 (4) 2048
2^11 - 2^6 => 1984
@gs4890 said:It will form up series(6.7 + 5.6.2 + 4.5.3 + 3.4.4 + 2.3.5 + 1.2.5) / 2 + 1 => 126bit lengthy qstn but can be done quickly with practice
Mere wale mai kya problem ho rhi?
@realslimshady said:s is a six digit number beginning with 1. if the digit is moved from the leftmost place to the rightmost, the number obtained is three times s. sum of digits of s is: a.21b.24 c.26 d.27with approach please
let the number be 1abcdef
with the above condition:
3*1abcdef=abcde1
300000+3abcde=10abcde+1
7abcde=300000-1=299999
abcde=42857
sum of digits od 1abcdef=1+4+2+8+5+7=27
@quant89 said:how many four digit numbers of the form PQRS are there with P is less than Q is less than R is less than S ?189, 199, 126, 303ans is 126ca an1 pls give an illustrative approach i couldn't unddderstand the approach given in the book???
we cant have 0 at the P place..
so selecting 4 no from 9 numbers and it can be arranged be in only 1 way
9c4 = 9.8.7.6/24 = 126
@quant89 said:@Torque024 yaar 126 hai ans TIME ke practice paper mein se hai..@gs4890 can u illustrate a bit..
selection of 4 numbers from 1 to 9 as we cannot use 0 here in 9C4 and in only one way we can arrange them
so , answer will be = 9C4
P.S :- nice username 
@rkshtsurana said:we cant have 0 at the P place..sp selecting 4 no from 9 numbers9c4 = 9.8.7.6/24 = 126
But we can have 0 in second place so 9*9*8*7 nai hona chaiyae?
@Torque024 said:But we can have 0 in second place so 9*9*8*7 nai hona chaiyae?
how ?
sequence should be like p
@Torque024 said:But we can have 0 in second place so 9*9*8*7 nai hona chaiyae?
P
no is PQRS. 0 kaha se aayega
Han aa gaya samaj, thanks. 😃
If F(x,y) is a remainder function signifying x mod y
f(8^643, 132)/(c-1) = f(52^60,31)
Then find the value of
f(((2^c+3^c+5^c+7^c+....+43^c+47^c)+2^5c)),2^3*3^2*5^1)
A)17
B)0
C)1
D)58
E)57
f(8^643, 132)/(c-1) = f(52^60,31)
Then find the value of
f(((2^c+3^c+5^c+7^c+....+43^c+47^c)+2^5c)),2^3*3^2*5^1)
A)17
B)0
C)1
D)58
E)57
@Ashmukh said:How many subsets of {1, 2, 3 €Ś 11} contain at least one even integer?(1) 1900 (2) 1964(3) 1984 (4) 2048
total cases = 2^11 - 1
when only odd = 2^6 - 1
when only odd = 2^6 - 1
required answer = 2^11 - 1 - 2^6 + 1 = 1984
@gautam22 said:2^c+3^c+5^c+7^c+....+43^c+47^c.....sir ye pehle 2 mein 1 ka gap hai fir 2 ka last mein 4 ka is series ko kya consider karein?
Prime number ki series
@gautam22 said:2^c+3^c+5^c+7^c+....+43^c+47^c.....sir ye pehle 2 mein 1 ka gap hai fir 2 ka last mein 4 ka is series ko kya consider karein?
PRIME NO HE SARE
@gautam22 said:is it by any chance 0?
Yeah, it is 0.
I can't get it, and have no solution.
Approach?
I can't get it, and have no solution.
Approach?
I chanced upon this question now. Floored! [image was removed because it was too large]
Options: 1, 2, 3, 4
Options: 1, 2, 3, 4