The given digits are 4, 4, 6, 6, 6, 7 and 0.
Totally we have 7 digits. So numbers greater than a million can be formed by using all the digits.
4 occurs twice, 6 occurs thrice while 0 and 7 once,
Therefore total number of arrangements = 7! / (3! x 2!) = 420 We have to avoid 0 in the starting place,
Number of ways in which 0 can be in the first place is 6! / 2! X 3! = 60
Hence the total number of ways = 420 – 60 = 360

Any divisor of 2⁶.3⁴.5⁷. is of the form 2^a.3^b.5^c. where 0≤ a ≤ 6, 0 ≤ b ≤ 4 & 0 ≤ c ≤ 7.
Thus the sum of the divisors of 2⁶.3⁴.5⁷.
is (1 + 2+…..+2⁵ + 2⁶)(1 + 3 + …. + 3⁴)(1 + 5 +….+ 5⁷) =
((2⁷-1)/(2-1))((3⁵-1)/(3-1))((5⁸-1)/(5-1))
=((2⁷-1)(3⁵-1)(5⁸-1)/2.4)

The total number of ways of forming the group of ten representatives is ²²C₁₀.

‘LOGARITHMS’ contains 10 different letters.

Total no. of seats,
= 1 grandfather + 5 sons and daughters + 8 grandchildren
= 14
The grandchildren can occupy the 4 seats on either side of the table in 4! = 24 ways.
The grandfather can occupy a seat in (5 – 1) = 4 ways (4 gaps between 5 sons and daughter).
And, the remaining seats can be occupied in 5! = 120 ways (5 seat for sons and daughter).
Hence total number of required ways,
= 8! × 480

Let S be the sample space

n(S) = 52C₃ = [( 52 x 51 x 50) / (3 x 2 x 1)] = 13600 / 6 = 22100

Let E = event of getting 3 kings out of 4.

n(E) = 4C₃ = [(4 x 3 x 2 x 1) / (3 x 2 x 1)] = 24/6 = 4

therefore, p(E) = [n(E) / n(S)]

==> p(E) = [ 4 / 22100]

==> p(E) = 1/5525

25! / 23! = 23! X 24 x 25 / ( 23!) = 24 x 25 = 600

(16 – 2)! × 2 = 14! × 2

Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.

The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.

The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.

Required number of numbers = (1 x 5 x 4) = 20.

Total 20 persons, 3 always seat together, 17 + 1 =18 delegates can be seated in (18 -1)! Ways = 17!
And now that three can be arranged in 3! Ways.
So, 17! 3! is the correct answer.

Maximum number of such different groups = ABC, ABD,ABE, BCE,BDE,CEA,DEA =7.

n the word, “MATERIAL” there are three vowels A, I, E.

The word ‘OPTICAL’ contains 7 different letters.

When the vowels OIA are always together, they can be supposed to form one letter.

Then, we have to arrange the letters PTCL (OIA).

Now, 5 letters can be arranged in 5! = 120 ways.

The vowels (OIA) can be arranged among themselves in 3! = 6 ways.

 Required number of ways = (120 x 6) = 720.

Number of letters in SMALL is 5, but there are 2 L?s Answer is 5! / 2! = 120 / 2 = 60

Six members can be selected from ten members in

Here total letters are 8,3 vowels and 5 consonants. Here 2 consonants can be chosen in ⁵C₂ ways
and these 2 consonants can be put it in 2! Ways. The remaining 6 letters can be arranged in 6! Ways.

The words beginning and ending letters with consonant = ⁵C₂ *2! *6! = 14400

To draw a line we need 2 points. From the given 8 points ,

we need to choose 2 points Number of ways in which this can be done is 18C2= (8 x 7 ) / 2 = 28

Therefore the required numbers of ways = 6*24 = 144

There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.

Let us mark these positions as under:

(1) (2) (3) (4) (5) (6)

Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5.

Number of ways of arranging the vowels = ³P₃ = 3! = 6.

Also, the 3 consonants can be arranged at the remaining 3 positions.

Number of ways of these arrangements = ³P₃ = 3! = 6.

Total number of ways = (6 x 6) = 36.

First Approach:

A number is divisible by 66 if it is divisible by 22 and 33.

Now the possible combinations I found are

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

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

(2,3,4)(2,3,4)

(3,2,4)(3,2,4)

(4,3,2)(4,3,2)

(3,4,2)(3,4,2)

(3,5,4)(3,5,4)

(5,3,4)(5,3,4)

total 88 ways.

Second Approach

Case1: unit digit can be filled in only two ways (2,4)(2,4) for nos (3,2,4)(3,2,4)

Tens digit can be filled in 22 ways

Hundred digit can be filled in 11 ways

the required number is 2*1*2=4 ways

Case2: unit digit can be filled in only one ways (2)(2) for nos (1,2,3)(1,2,3)

Tens digit can be filled in 11 ways

Hundred digit can be filled in 22 ways

the required number is 2*1*1=2 ways

Case3: unit digit can be filled in only one ways (4)(4) for nos (3,4,5)(3,4,5)

Tens digit can be filled in 11 ways

Hundred digit can be filled in 22 ways

the required number is 2*1*1=2 ways

So, Total ways=8

DIRECTOR

Total Letters = 8

Vowels=3

Consonants= 5

Considering all vowels together as 1 group and all consonants together as another group.

We have 2 groups one of vowels and other of consonants.

So, Total permutation of 2 groups = 2!=2 ways.

Considering group of vowels permutations possible in the group of vowels =3!=6 ways.

Considering group of consonants permutations possible in the group of consonants =5!/(2!)=5×4×3=60 ways

Hence ,Total permutations =2×6×60=720

In the word ‘MATHEMATICS’, we’ll consider all the vowels AEAI together as one letter.
Thus, we have MTHMTCS (AEAI).
Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice
Number of ways of arranging these letters =8! / ((2!)(2!))= 10080.

Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.
Number of ways of arranging these letters =4! / 2!= 12.

Required number of words = (10080 x 12) = 120960

The word ‘OPTICAL’ contains 77 different letters. When the vowels OIA are always together, they can be supposed to form one letter. Then, we have to arrange the letters PTCL (OIA). Now, 55 letters can be arranged in 5!=1205!=120ways. The vowels (OIA) can be arranged among themselves in 3!=63!=6 ways. Required number of ways =(120∗6)=720=(120∗6)=720.

The number of arrangement in which A and B are not together

Task 2 must be assigned to P₄ or P₃ =2C₁

For task 1 only 2 persons i.e.., P₃ or P₄and P₅ = 2C₁

Task 3 only 3 person available =3C₁

For task 4 only 2 persons available =2C₁

For task 5 only 1 person left = 1

Therefore, total number of ways

= 2C₁ x 2C₁x 3C₁x 2C₁ x1 = 2x2x3x2x1 =24

This type of question becomes very easy when we assume three colour are red(R) blue(B) and Green(G).

As there are six players, So total ways in which they can be arranged = 6!ways =720.

A number of ways in which Ajeet and Mukherjee are together = 5!x2 = 240.

Therefore, Number of ways when they don’t remain together = 720 -240 =480.

The number of ways of selecting three men, two women and three children is: