We must have
xW + xR + xB = 10.
Required number of ways
= Number of non-negative integral solutions of
xW + xR + xB = 10
= 3 + 10 – 1C 10 =^12 C 10 =^12 C 2
- (b) : Person who has to sit between the brothers can
be selected in^8 C 1 ways.
Thus, total ways = 8·(7!) · (2!) = (8!) · (2!) - (a) : Total number of books = a + 2b + 3c + d
Total number of ways in which these books can be
arranged in a shelf (in same row)
= +++
⋅()!
!( !) ( !)abcd
ab c23
23- (b) : That means formed number can be atmost of
three digits.
Total number of one digit numbers = 6
Total number of two digit numbers = 5. 5 = 25
Total number of three digit numbers= 5. 5. 4 = 100
Thus, total number of such numbers = 131 - (a) : Let the number is x 1 x 2 x 3 x 4 x 5 x 6.
We m u s t h a v e
x 1 < x 2 < x 3 < x 4 < x 5 < x 6.
Clearly no digit can be zero.
Thus total number of such numbers =^9 C 6 =^9 C 3. - (a) : Alphabetical order of these letters is B, E, K, R,
U.
Total words starting with B = 4! = 24
Total words starting with E = 4! = 24
Total words starting with KB = 3! = 6
Total words starting with KE = 3! = 6
Total words starting with KR = 3! = 6
Next word will be KUBER = 24 + 24 + 18 + 1 = 67 - (b) : Let there were ‘n’ players in the beginning.
Total number of games to be played was equal to nC 2
and each player would have played (n – 1) games. Thus
nC 2 – ((n – 1) + (n – 1) – 1) + 6 = 84
⇒ n^2 – 5n – 150 = 0
⇒ n = 15
- (a) : All strips are of different colours, then number
of flags = 3! =6
When two strips are of same colour, then
number of flags =^31321
2
CC. !. = 18∴ Total flags = 6 + 18 = 24 = 4!
- (d) : Total number of triangles
=^3 C 1. 4 C 1. 5 C 1 +^3 C 2 (^4 C 1 +^5 C 1 )
+^4 C 2 (^3 C 1 +^5 C 1 ) +^5 C 2 (^3 C 1 +^4 C 1 )
= 3. 4 .5 + 3.9 + 6.8 + 10.7 = 205
- (d) : There will be (n 1 + 1) gaps created by n 1 men.
Now women have to be seated only in these gaps.
Thus number of such sitting arrangements
=n^1 +^1 Cnnn 2 ⋅⋅ 12 !! - (b) : When seats are numbered, circular permutation
is same as linear permutation. Thus total number of
sitting arrangements is equal to nPn 1. - (a) : We can think of three groups. One consisting
of three boys of class X, other consisting of 4 boys of
class XI and last one consisting of 5 boys of class XII.
These groups can be arranged in 3! ways and boys in
these groups can be further arranged in 3!, 4! and 5!
ways respectively.
Thus total ways = (3!)^2 (4!) (5!) - (a) : 1M, 4I, 2P and 4S. Total gaps created by letter
1M, 4I and 2P is 8.
Thus total arrangements=^84 ⋅^7 =
24
C! 7350
!!- (b) : First of all two men can be selected in^9 C 2 ways.
Thereafter 2 women can selected in^7 C 2 ways (as wives
of selected men are not be selected). And finally they
can be paired up in^2 C 1 ways.
Thus total ways =^9 C 2 ·^7 C 2 ·^2 C 1. - (a) : Choices available to the candidates are; 3
questions from first group and 4 from another or 4
questions for first group and 3 questions from another.
Thus total ways = 2 ·^5 C 3 ·^5 C 4 - (c) : x 1 < x 2 ≤ x 3 < x 4 < x 5 ≤ x 6
It will give rise to following four cases :
(i) x 1 < x 2 < x 3 < x 4 < x 5 < x 6 → 9 C 6 ways
(ii) x 1 < x 2 = x 3 < x 4 < x 5 < x 6 → 9 C 5 ways
(iii) x 1 < x 2 < x 3 < x 4 < x 5 = x 6 → 9 C 5 ways
(iv) x 1 < x 2 = x 3 < x 4 < x 5 = x 6 → 9 C 4 ways
Thus total number of such numbers
=^9 C 6 +^9 C 5 +^9 C 5 +^9 C 4 =^10 C 6 +^10 C 5 =^11 C 6