1. Number of selections of atleast 10 articles from 19 different articles is.

A) 2^19 B) 2^18 C) 2^19 - 1 D) 2^18 - 1 E) None

Hello abhi 07

Number of ways for selection of at least 10 articles from 19 different articles is

19c_{0} +^{19}c_{1} +^{19}c_{2 }+^{19}c_{3}+^{19}c_{4+.....}+^{19}c_{10}=E

Also, ^{19}c_{1} +^{19}c_{2 }+^{19}c_{3}+^{19}c_{4+.....}+^{19}c_{19 }= 2^{19}

And ^{19}c_{0} =^{19}c_{19 }, ^{19}c_{2}=^{19}c_{17+.....}

Thus E+E=2^{19} and E=2^{18}

Please don't generate new topic,

Post new questions on the topic which already existed.

There were 90 questions in an exam. If 1 mark was awarded for every correct answer and 1/3^{rd} mark was deducted for every wrong answer, how many different net scores were possible in the exam?

How many different size squares are there in a square grid of five by seven?

Hello Reeshabh,

from 1 x 1 to 5 x 5

5 different size squares .

What is the sum of all the four digit numbers made using the digits 1,2,3,4 (with as well as without repetition)?

Hello Richa ,

1234 = 10³ • 1 + 10² • 2 + 10 • 3 + 4

10³ • 1 will occur altogether in 3! ways similarly each of 10³•2 , 10³•3 , 10³•4 will occur in 3! ways .

10³•1 + 10²•2 + 10•3+ 4

10³•2 + 10²•3 + 10•4 + 1

10³•3 + 10²•4 + 10•1 + 2

10³ •4 + 10²•3 + 10•2 + 1

Required sum :

3! × ( 1 + 2 + 3 + 4 ) + (10³+ 10² + 10 +1)

Hence , 6 × ( 1+2+3+4) × 1111 = 66660

If repition allowed :

_ _ _ _

Keep one digit fixed . We are left with 3 blanks and 4 digits

So fill them in 4 × 4 × 4 ways

Hence , (1 + 2 + 3 + 4) × 4³ × 1111 = 711040

how many pairs of divisors of 21600 will have h.c.f 6..?

Hello Aniket ,

6k and 6n are the factors , where k and n are coprime to each other .

=> k,n are the coprime factors of 3600.

3600 = 2^{4} 3^{2 }5^{2}

k = 2^{a1} 3^{b1} 5^{c1 }

n = 2^{a2} 3^{b2} 5^{c2}

one of a_{1} , a_{2} has to be 0

=> ( 5 x 5) – ( 4 x 4) = 9 cases

one of b_{1} , b_{2} has to be 0

=> ( 3 x 3 ) – ( 2 x 2 )= 5 cases

one of c_{1} , c_{2} has to be 0

=> ( 3 x 3 ) – ( 2 x 2 )= 5 cases

Hence , total number of ordered pair solution : 9 x 5 x 5 – 1 = 224 cases

Unordered pairs 224/2 = 112

Please share the solution to the problem below:

In all the words formed by the letters of the word RAINBOW are arranged in a dictionary form, then what is the position of the word RAINBOW in that dictionary

A) 3136

B) 3361

C) 3631

D) 1363

Hello Richa,

Arrange letters in alphabetical order: ie, A,B,I,N,O,R,W.

Now words starting with A/B/I/N/O will come

ahead of the words starting with R.

Words starting from A :

A _ _ _ _ _ _

you can arrange remaining letters in 6! ways .

6! = 720

Same with B, I, N, O. That makes it 720 x 5 = 3600 words

3601st Word will start with R followed by A,

followed by B. The remaining 4 letters can be arranged in 4! = 24

ways. Then Next would be RAIB => (3!) = 6 ways. Then next would

be RAINBOW. Add and get the answer.

You can see words starting with A , B , I , N and O make up 3600 words . By options if you go only option (C) is more than 3600 . Hence 3631 will be the answer .

In how many ways can you select exactly 7 letters from 3A, 4B, 2C and 1D?

Hello Samyak,

Selecting 7 out of given 10 letters is same as not selecting 3letters out of 10

Hence ,

Three case :

(i)All same ( ppp)

2 ways : 3A's or 3B's

(ii) 2 same (ppq)

3c2 × 3 = 9 ways

(iii) three distinct ( pqr)

4C3 = 4 ways

Total : 9 + 4 + 2 = 15 ways

HI SIR, could you please explain this question? P.S- I did not understand it from the solution given in the mock.

Two cards are randomly selected from a well-shuffled deck of 52 playing cards. Find the probability that one of them is a Queen and the other is a black card.

Hello Richa

A standard deck of card has 52 cards.

Equal number of cards(13) of 4 different suits :

Spade - Black

Diamond -Red

Heart - Red

Club - Black

We have to select a queen and a black card :

there are 26 black cards and 4 queens ( 2 red , 2 black )

Two cases :

(i) Red Queen and a Black Card

(ii) Black Queen and A Black Card

For case (i) -

select a red queen from 2 red queens in 2C1 ways and a black card in 26C1ways

Hence , 2 x 26 = 52 ways

For case (ii)

total number of ways to select two black cards = 26C2

Number of ways to select two black non-queen cards = 24C2

Hence, the number of ways in which at least one black queen ( at most 2) is chosen

26C2 - 24C2 = 49

Probability = ( 49 + 52)/(52C2)

Sir, why can't in case (ii) we simply do 2C1 (for one black queen) and 25C1 (selecting one card from the remaining 25 black cards. This way also we will get at least one (and at most 2 queens).

But this gives the answer as 50 instead of 49. So, what is wrong with this method?

If you do this, there is one case where you have chosen BOTH the queens in your selection. As per the question, you should have ONE queen and ONE black card.

Hello Diksha,

Number of whole number solutions for

a + b + c + .......( r terms ) = N

=> ( N + r -1) C ( r -1)

When both ( chocolates and boxes ) identical make cases :

Distributing 1 , 2, 3 ,4 , ......., 9 , 10 chocolates we are left with 57 - 55 = 2 chocolates .

1. ( 1 , 2, 3, 4, 5, 6, 7, 8, 9 + 1 , 10 + 1 )

2. ( 1 , 2, 3, 4, 5 , 6, 7 ,8 . 9+2 , 10)

3. ( 1 , 2,3, 4, 5 , 6, 7 , 8 , 9 , 10+2 )

3 ways .

how many pairs of x and y are possible if

x^{2}-5y^{2}=1232

Every perfect square can be written as 5k, 5k+1 or 5k + 4 form.

If x^{2} is in the form of 5k then x^{2} – 5y^{2} will be a multiple of 5.

If x^{2} is in the form of 5k+1 then x^{2}– 5y^{2} will be 1 more than a multiple of 5.

If If x^{2} is in the form of 5k+4 then x^{2}– 5y^{2} will be 4 more than a multiple of 5.

Now we can see 1232 gives 2 as a remainder when divided by 5,

Hence, x^{2} – 5y^{2} = 1232 will have no integer solution.

Four circular tables are arranged such that their centers from a square. If each table can accommodate 4 persons, in how many ways can 16 people be selected across such an arrangement?

Sir, I have a doubt in this question. I agree that for the 1st person, all the tables are alike so he can select any one table in 1 way. After that, why are there 4 different ways for him to be seated at that table? Aren't the chairs on that particular table too alike for him? So shouldn't the ways of selecting the chair be 1 instead of 4? Please clarify.

A box contains 5 chips, numbered from 1,2,3,4 and 5. Chips are drawn randomly one at a time without replacement until the sum of the values drawn exceeds 4. What is the probability that 3 draws are required?

4 cases :

( 1 , 2 ) , ( 2 ,1) , (1,3) and (3,1)

probability of getting 1 is : 1/5 and probability of getting 2 from the remaining 4 chips is 1/4

Hence , 1/4 x 1/5 = 1/20

Each of these 4 cases has 1/4 x 1/5 chance . Hence , (1/20) x 4 = 1/5