প্রতিটি আংটির জন্য, সেটি পরার জন্য 4টি ভিন্ন আঙুল উপলব্ধ আছে। যেহেতু আংটিগুলো ভিন্ন এবং একটি আঙুলে একাধিক আংটি পরা যেতে পারে, তাই প্রতিটি আংটির জন্য স্বাধীনভাবে আঙুল বেছে নেওয়া যেতে পারে।

প্রথম আংটির জন্য, পরার উপায় = 4টি আঙুল
দ্বিতীয় আংটির জন্য, পরার উপায় = 4টি আঙুল
তৃতীয় আংটির জন্য, পরার উপায় = 4টি আঙুল
চতুর্থ আংটির জন্য, পরার উপায় = 4টি আঙুল
পঞ্চম আংটির জন্য, পরার উপায় = 4টি আঙুল

সুতরাং, মোট যত উপায়ে 5টি ভিন্ন আংটি 4টি আঙুলে পরা যেতে পারে তা হলো এই উপায়গুলোর গুণফল = 4×4×4×4×4 = 1024
অতএব, পাঁচ ভিন্ন আংটি এক হাতের চারটি আঙুলে 1024 উপায়ে পরা যেতে পারে।

১ থেকে ১৫ পর্যন্ত মোট ৬টি মৌলিক সংখ্যা রয়েছে
সেগুলো হলো ২, ৩, ৫, ৭, ১১, ১৩

একটি মৌলিক সংখ্যা নির্বাচন করার সম্ভাবনা হলো = ৬/১৫ = ২/৫ = ০.৪

প্রথম ব্যক্তি ৯ জনের সাথে হাত মেলাবে (১০ - ১ = ৯)।
দ্বিতীয় ব্যক্তি আর ৮ জনের সাথে হাত মেলাবে (যেহেতু প্রথম ব্যক্তির সাথে তার হ্যান্ডশেক গণনা হয়ে গেছে)।
তৃতীয় ব্যক্তি আর ৭ জনের সাথে হাত মেলাবে।
...
নবম ব্যক্তি আর ১ জনের সাথে হাত মেলাবে।
দশম ব্যক্তির সাথে সকলের হ্যান্ডশেক আগেই গণনা হয়ে গেছে।

সুতরাং, মোট হ্যান্ডশেকের সংখ্যা = ৯ + ৮ + ৭ + ৬ + ৫ + ৪ + ৩ + ২ + ১ = ৪৫।

প্রদত্ত অঙ্ক মোট ৬টি যার মধ্যে '৩' দুইটি '৪' তিনটি।

নির্ণেয় ছয় অঙ্কবিশিষ্ট মোট গঠিত সংখ্যা
= 6!/(2! × 3!) টি
= (6×5×4×3×2× 1)/(2×1×3×2×1)টি
= 60টি

2¹² = 4096 

As no two boys are to be together, we need the presence of a girl between every two boys.

The seven girls can be arranged in 7! Ways.

_G_G_G_G_G_G_G_

The five boys can be arranged in 8 available gaps in ⁸P₅ ways.

∴ the total number of arrangements = 7! × ⁸p₅

We see all letters of word FACETIOUS are different.

Take any three letters as first three letters. This can be done in 9C3 ways, i.e., 84 ways.

Now, for these first three letters, there is only one arrangement possible that is when they are in dictionary order.

Remaining 6 letters can be arranged in 6! Ways.

∴ Total number of ways = 84 × 6! = 60480

Exactly two of them are hearts and exactly two of them are kings. This is possible in two ways.

Case i) Out of the 4 cards, no card is a king of hearts.

⇒ Two hearts can be chosen in ¹²c₂ ways, i.e., 66 ways.

And, two kings can be chosen in ³ c₂ ways, i.e., 3 ways.

∴ Total numbers of ways in this case = 66 × 3 = 198

Case ii) Out of the 4 cards, one card is a king of hearts.

Now, we have one card fixed. Out of remaining 3 cards, one should be a heart and one should be king. The third card can be any card that is neither king nor heart.

∴ Total numbers of ways in this case = 12 × 3 × (52 – 13 – 3) = 1296

∴ Total number of ways in which it can be done = 198 + 1296 = 1494

In a four digit number, first digit is 9 and last digit is smallest among 4 digits. Sum of middle two digits is even.

⇒ Between first and last digits, there is a difference of at least 3 digits.

It may come into mind that a difference of 2 digits can suffice, but that would lead to sum of middle two digits being odd (sum of two consecutive digits). But we need the sum to be even.

Combinations for first and last digits can be (9, 0), (9, 1), (9, 2), (9, 3), (9, 4), (9, 5).

Between 0 and 9, there are 4 odd digits and 4 even digits. Two can be chosen such that their sum is even, if both are odd or both are even. This can be done in 4 × 3 + 4 × 3 = 24 ways.

Similarly, for 1 and 9, there can be 3 × 2 + 4 × 3 = 18 ways.

For 2 and 9, there can be 12 ways. For 3 and 9, there can 8 ways. For 4 and 9, there can be 4 ways. For 5 and 9, there can be 1 way.

∴ Total number of 4 digit numbers possible = 24 + 18 + 12 + 8 + 4 + 1 = 67

The given word is COUSIN. If vowels are arranged in dictionary order, then I will come before O and O will come before U.

There are 6 letters in this word which need to be arranged.

Let’s first arrange the vowels. They can be placed in any 3 of the 6 places.

Number of ways of choosing these places = ⁶c₃

There’s only one way of arranging the vowels in the selected places, and that is the alphabetical order.

Now, the remaining three letters, i.e. C, S and N are to be placed in remaining 3 places. This can be done in 3! Ways.

∴ Total number of possible arrangements = ⁶c₃ × 3! = 6 × 5 × 4 = 120

Each pencil can be put in one of three bags.

⇒ Number of ways in which this can be done = 3 × 3 × 3 × 3 × 3 × 3 = 729

Out of these, there can be some cases when one or more bags are empty, and when all bags have same number of pencils.

Case i) one bag is empty

Empty bag can be selected in ³c₁ ways i.e. 3 ways

Now, each pencil can be put in one of two ways.

Number of ways in which this can be done = 3 × (2 × 2 × 2 × 2 × 2 × 2) = 192

Case ii) two bags are empty

All pencils will be in one of three bags. This can be done in only 3 ways.

Case iii) three bags are empty. This case is not possible

Case iv) all bags have same number of pencils, i.e., 2 pencils.

⇒ We have to select two pencils for each bag.

Number of ways in which this can be done = ⁶c₂ × ⁴c₂ × ²c₂ = 15 × 6 × 1 = 90

∴ Total number of ways in which this can be done = 729 – 192 – 3 – 90 = 444

First we will find Quantity A,

Quantity A:

10 × ⁿc₂ = 3 × n+ ¹c₃

∵ ⁿc_r = \(\frac{{n!}}{{\left( {n - r} \right)!\left( {r!} \right)}}\)

\(\Rightarrow \;10 \times \frac{{n\left( {n - 1} \right)}}{{1 \times 2}}\; = \;3 \times \frac{{\left( {n + 1} \right)\left( {n - 1} \right)\left( {n} \right)}}{{1 \times 2 \times 3}}\)

 ⇒ 10 = n + 1

⇒ n = 9

∴ n = 9

Now,

Quantity B:

ⁿC₄ = ⁿC₆

(∵ ⁿc_r = ⁿc_s then r = s or r+s = n)

⇒ n = 4 + 6 = 10

∴ n = 10

∴ Quantity B > Quantity A

Let the total number of person be N

∵ Total number of hand-shakes is 55

∵ For a hand shake we require two people, total number of handshake is NC2

∴ ⁿC₂ = 55

∴ N = 11 persons

A million is 1000000(first seven digit’s number). So, we need to find how many numbers of less than 7 digits can be formed using the digits 0, 7 and 8.

This can be done as there are 2 possibilities at this first digit ⇒ 7 and 8

At all other places, any of the three numbers can be present. So there are three possibilities for all other digit’s places, therefore possible numbers

Number of 1 digit numbers = 2

Number of 2 digit numbers = 2 × 3 = 6

Number of 3 digit numbers = 2 × 3 × 3 = 18

Number of 4 digit numbers = 2 × 3 × 3 × 3 = 54

Number of 5 digit numbers = 2 × 3 × 3 × 3 × 3 = 162

Number of 6 digit numbers = 2 × 3 × 3 × 3 × 3 × 3 = 486

Total number of such numbers = 2 + 6 + 18 + 54 + 162 + 486 = 728

First of all we will make circular permutations for all the flowers of a particular colour.

We know that, for permutation of n objects in a circle is = (n - 1)!

So, for 10 flowers of same colour, the permutations in 10 places = (10 - 1)! = 9!

Now, for permutation of the remaining ten flowers of same colour in between the previously placed flowers = 10!

Now, since this is a necklace hence the number of permutations are divided by 2.

Hence, the required answer = (9! × 10!)/2 = 5(9!)²

This problem can be solved using the M-N-P rule.

Rule – If there are three things to do and there are M ways to do the first thing, N ways to do the second thing and P ways to do the third thing then there will be M × N × P ways to do all the three things together.

Thus, to travel from Allahabad to Kolkata via Lucknow

There are M = 5 routes to travel from Allahabad to Lucknow

There are N = 4 routes to travel from Lucknow to Kolkata

Then, M × N = 5 × 4 = 20 ways to travel from Allahabad to Kolkata via Lucknow.

If we put 1 marble in the first box and the rest in the other, then the number of ways it can be done

= ⁶C₁ × ⁵C₅ = 6 × 1 = 6

If we put 2 marble in the first box and the rest in the other, then the number of ways it can be done

= ⁶C₂ × ⁴C₄  = 15 × 1 = 15

If we put 3 marble in the first box and the rest in the other, then the number of ways it can be done

= ⁶C₃ × ³C₃ = 20 × 1 = 20

If we put 4 marble in the first box and the rest in the other, then the number of ways it can be done

= ⁶C₄ × ²C₂  = 15 × 1 = 15

If we put 5 marble in the first box and the rest in the other, then the number of ways it can be done

= ⁶C₅ × ¹C₁ = 6 × 1 = 6

The number of ways that 6 different marbles can be put in boxes of different sizes so that no box remain empty is = 6 + 15 + 20 + 15 + 6 = 62

When L and T are fixed as first and last letters of the word, then we have only 6 letters to arrange.

Hence, required number of words  \(= {{!6} \over {!2 \times !2 \times !2}} = {{720} \over 8} = 90\)\({!2 × !2 × !2 ⇒ because M, A I occurs two times}.\)

Total number of letters in ARRAGEET = 8

In the given word there are 2 “R”, 2 “A” and 2 “E”.

As we know that number of ways of arranging a letters out of which r are of same type, m are of same type, n are of same type and p are of same type = \(\frac{{{\text{a}}!}}{{{\text{r}}!{\text{m}}!{\text{n}}!{\text{p}}!}}\)

So total number of ways = \(\frac{{8!}}{{2!2!2!}} = 5040\)

Consider the 3 girls as one unit. Then we have 4 + 1 = 5 people. They can be arranged in 5! = 120 ways. The 3 girls among themselves can be arranged in 3! = 6 ways.

The number of required arrangements = 120 × 6 = 720 ways.

Hence, the answer is 720,

In a simultaneous throw of two dice –

Number of possible outcomes n(s) = 6 × 6 = 36

Number of favourable events = n(E) = {(1, 6); (2, 5), (3, 4), (4, 3), (5, 2), (6,1)}

⇒ n(E) = 6

∴ Required probability of getting a total of 7 –

\(P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}}\)

\(= \frac{6}{{36}} = \frac{1}{6}\)

The first letter can be chosen in 26 ways. Similarly, second letter can be chosen in 26 ways.

Hence, the first 2 letters can be chosen in 26 × 26 ways.

Also, the third letter can be chosen in 26 ways.

Hence, the first 3 letters can be chosen in 26 × 26 × 26 ways.

But the fourth letter should be similar to second letter. This can be done in just 1 way.

Similarly, fifth letter should be similar to first letter. This can be done only in 1 way.

Hence, number of words possible = 26 × 26 × 26 × 1 × 1

= 17576

Case – I         

Committee contains 2 women and 1 men.

∴ No. of ways of forming this committee = (number of ways of choosing 2 women out of 3) × (number of ways of choosing 1 man out of 5)

= ³C₂ ⋅ ⁵C₁ =3 × 5 = 15

Case – II        

Committee contain 3 women and 0 men

No. of ways of forming this committee = number of ways of choosing 3 women out of 3.

= ³C₃  =1

∴ Total number of ways = 15 + 1 = 16

First digit can be 1, 2, 3 only. (four digit number)

Hence, 3 × 3 × 2 × 1 = 18 ways (with no digit repetition)

We know that:

Formula:

\({}_{}^n{P_r} = \frac{{!n}}{{!\left( {n - r} \right)}}\)

∴⁶P₂ × ⁵P₃ =  \(\frac{{!6}}{{!\left( {6 - 2} \right)}} \times \frac{{!5}}{{!\left( {5 - 3} \right)}}\)

⇒⁶P₂ × ⁵P₃ \(= \frac{{!6}}{{!4}} \times \frac{{!5}}{{!2}}\)

⇒⁶P₂ × ⁵P₃ = \(\frac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{4 \times 3 \times 2 \times 1}} \times \frac{{5 \times 4 \times 3 \times 2 \times 1}}{2}\)

⇒⁶P₂ × ⁵P₃ = 30 × 60

⇒⁶P₂ × ⁵P₃  = 1800

 

x=√17956=134