Total number of balls = 2 + 3 + 4 = 9
Total number of ways 3 balls can be drawn from 9 = 9C3
No green ball is drawn = 9 - 3 = 6 = 6C3
Required number of ways if atleast one green ball is to be included = Total number of ways - No green ball is drawn
= 9C3 - 6C3
= 9x8x7/3x2 - 6x5x4/3x2
= 84 - 20
= 64 ways.
Given word is SCOOTY
ATQ,
Except S & Y number of letters are 4(C 2O's T)
Hence, required number of arrangements = 4!/2! x 2! = 4!
= 4 x 3 x 2
= 24 ways.
Let 'Y' be the youngest player.
The last song can be sung by any of the remaining 3 players. The first 3 players can sing the song in (3!) ways.
The required number of ways = 3(3!) = 4320.
Number of mens = 8
Number of womens = 5
Different ways could couples be picked = = 9 x 6 = 54 ways.
Total Combination of getting a card from 52 cards =
Because there is no replacement, so number of cards after getting first card= 51
Now, Combination of getting an another card=
Total combination of getting 2 cards from 52 cards without replacement= (
)
There are total 4 Ace in stack. Combination of getting 1 Ace is =
Because there is no replacement, So number of cards after getting first Ace = 3
Combination of getting an another Ace =
Total Combination of getting 2 Ace without replacement=
Now,Probability of getting 2 cards which are Ace = = 1/221.
Number of positive integral solutions = = = 190
7 × 5 = 35
35 ? 1 = 34
As A1 speaks always after A2, they can speak only in 1st to 9th places and
A2 can speak in 2nd to 10 the places only when A1 speaks in 1st place
A2 can speak in 9 places the remaining
A3, A4, A5,...A10 has no restriction. So, they can speak in 9.8! ways. i.e
when A2 speaks in the first place, the number of ways they can speak is 9.8!.
When A2 speaks in second place, the number of ways they can speak is 8.8!.
When A2 speaks in third place, the number of ways they can speak is 7.8!. When A2 speaks in the ninth place, the number of ways they can speak is 1.8!
Therefore,Total Number of ways they can speak = (9+8+7+6+5+4+3+2+1) 8! = = 10!/2
CAPITAL = 7
Vowels = 3 (A, I, A)
Consonants = (C, P, T, L)
5 letters which can be arranged in
Vowels A,I =
No.of arrangements = 5! x =360
There are 7 letters in the word ?Bengali of these 3 are vowels and 4 consonants.
Considering vowels a, e, i as one letter, we can arrange 4+1 letters in 5! ways in each of which vowels are together. These 3 vowels can be arranged among themselves in 3! ways.
Total number of words = 5! x 3!= 120 x 6 = 720
4 novels can be selected out of 5 in
ways.
2 biographies can be selected out of 4 in
ways.
Number of ways of arranging novels and biographies =
= 30
After selecting any 6 books (4 novels and 2 biographies) in one of the 30 ways, they can be arranged on the shelf in 6! = 720 ways.
By the Counting Principle, the total number of arrangements = 30 x 720 = 21600
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