Given total 16 Red roses and 14 White roses = 30 roses
Four flowers have to be selected from 30 i.e, = 27405 Ways
Now, atleast one Red rose is selected i.e, 27405(total) - 1(all four are white roses) = 27404 ways.
Given total number of students in the class = 21
So each student will have 20 greeting cards to be send or receive (21 - 1(himself))
Therefore, the total number of greeting cards exchanged by the students = 20 x 21 = 420.
Number of ways of choosing 2 black pens from 5 black pens in ways.
Number of ways of choosing 2 white pens from 3 white pens in ways.
Number of ways of choosing 2 red pens from 4 red pens in ways.
By the Counting Principle, 2 black pens, 2 white pens, and 2 red pens can be chosen in 10 x 3 x 6 =180 ways.
The students should sit in between two teachers. There are 7 gaps in between teachers when they sit in a roundtable. This can be done in ways. 7 teachers can sit in (7-1)! ways.
Required no.of ways is =
ST candidates vacancies can be filled by ways = 10
Remaining vacancies are 5 that are to be filled by 12
=> = (12x11x10x9x8)/(5x4x3x2x1) = 792
Total number of filling the vacancies = 10 x 792 = 7920
Number of cards in a pack of cards = 52
Number of black cards = 26
Number of king cards = 4 (2 Red, 2 Black)
Required, the probability that if a card is drawn either card is black or a king =
The number of letters in the given word RITUAL = 6
Then,
Required number of different ways can the letters of the word 'RITUAL' be arranged = 6!
=> 6 x 5 x 4 x 3 x 2 x 1 = 720
A Committee of 5 persons is to be formed from 6 gentlemen and 4 ladies by taking.
(i) 1 lady out of 4 and 4 gentlemen out of 6
(ii) 2 ladies out of 4 and 3 gentlemen out of 6
(iii) 3 ladies out of 4 and 2 gentlemen out of 6
(iv) 4 ladies out of 4 and 1 gentlemen out of 6
In case I the number of ways = = 4 x 15 = 60
In case II the number of ways = = 6 x 20 = 120
In case III the number of ways = = 4 x 15 = 60
In case IV the number of ways = = 1 x 6 = 6
Hence, the required number of ways = 60 + 120 + 60 + 6 = 246
DESIGN = 6 letters
No consonants appear at either of the two ends.
= = 2 x 4 x 3 x 2 x 1= 48
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 = = 3! = 6.
Also, the 3 consonants can be arranged at the remaining 3 positions.
Number of ways of these arrangements = = 3! = 6.
Total number of ways = (6 x 6) = 36.
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