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

**The number of ways of distributing 15 identical balloons, 6 identical pencils and 3 identical erasers among 3 children, such that each child gets at least four balloons and one pencil, is**

ANSWER 1

Given that, there are 15 identical balloons, 6 identical pencils, and 3 identical erasers.

We need to distribute these items among 3 children, such that each child gets at least four balloons and one pencil.

So, every person gets four balloons and one pencil. After distribution, we have 3 balloons, 3 pencils, and 3 erasers remaining.

So, let’s first arrange the balloons in all possible combinations.

(3, 0, 0) , (2, 1, 0) , (1, 1, 1)

We can arrange (3, 0, 0) in three ways.

We can arrange (2, 1, 0) in six ways.

We can arrange (1, 1, 1) in only one way.

So, we can arrange balloons among three children in 10 ways.

Likewise, we can also arrange 3 pencils in 10 ways and 3 erasers in 10 ways.

Therefore, the total number of ways is 10 x 10 x 10 = 1000 ways

**Q.)****The number of ways of distributing 15 identical balloons, 6 identical pencils and 3 identical erasers among 3 children, such that each child gets at least four balloons and one pencil, is**

ANSWER 1

Given that, there are 15 identical balloons, 6 identical pencils, and 3 identical erasers.

We need to distribute these items among 3 children, such that each child gets at least four balloons and one pencil.

So, every person gets four balloons and one pencil. After distribution, we have 3 balloons, 3 pencils, and 3 erasers remaining.

So, let’s first arrange the balloons in all possible combinations.

(3, 0, 0) , (2, 1, 0) , (1, 1, 1)

We can arrange (3, 0, 0) in three ways.

We can arrange (2, 1, 0) in six ways.

We can arrange (1, 1, 1) in only one way.

So, we can arrange balloons among three children in 10 ways.

Likewise, we can also arrange 3 pencils in 10 ways and 3 erasers in 10 ways.

Therefore, the total number of ways is 10 x 10 x 10 = 1000 ways

Q1.If we rolled a dice 3 times. In how many ways we can get a sum of 8?

Q2.

In how many ways we can pick 4 numbers from first 20 natural numbers such that there is at least a gap of one number between any two picked numbers?

ANSWER 1

(6,1,1) = 3 ways

(5,2,1) = 6 ways

(4,2,2) = 3 ways

(4,3,1) = 6 ways

(3,3,2) = 3 ways

The total no. of ways = 21 ways

ANSWER 2

Numbers before we pick first number = A

Numbers between first and second number = B

Numbers between second and third number = C

Numbers after we pick fourth number = D

ATQ,

A+B+C+D=16 [A≥0, B≥1, C≥1, D≥0]

A+B+C+D=14 [A≥0, B≥0, C≥0, D≥0]

NO OF WAYS= 14C2 = 91 WAYS

Q1. Five racquets needs to be placed in three boxes. Each box can hold all the five racquets. In how many ways can the racquets be placed in the boxes so that no box can be empty if all racquets are different but all boxes are identical?

A) 24

B) 25

C) 27

D) 26

Solution: B

Let the 5 racquets be a, b, c, d, e

3 Boxes are Identical

Since the boxes are identical 3 in First box, 1 in Second and 1 in third is not different from 1 in first, 3 in second and 1 in third.

But there is a catch in the above table.

Imagine this process, after selecting 3 out of a, b, c, d, e

Let's say we select d, e, c. a and b have to be put in two identical boxes.

This can be done only in one way.

In second scenario let's assume a and b in First box and c and d in second box and e in third box. But the possibility will be same when c and d in First box and a and b in second box.

So, it is ways, In first scenario 5C3 = 10 ways

Totally 25 ways.

Q1. There are 12 copies of Beetles CDs, 7 copies of Pink Floyd CDs, 3 different CDs of Michael Jackson, and 2 different CDs of Madonna. Find the number of ways in which one or more than one CD can be selected?

a) 3388

b) 3376

c) 3366

d) 3327

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?

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.

Sir could you please explain the solution after a+b+c=15

Hello Diksha,

Number of whole number solutions for

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

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