Answer:
Step-by-step explanation:
a) No two cards should be of the same colour. So for each of the seven cards we have to choose them from different colours. And for each colour we have 7 cards.
So for the required selection: We first select a card from one particular colour in 7C1 = 7 ways. Then we select the second card for the hand from one of the rest of the colours in 7C1 = 7 ways again as the other colour also has 7 cards. And so on.
So no. of ways in which we have a hand of no two cards to be same = 7 x 7 x 7 x 7 x 7 x 7 x 7 = 7^7 = 8,23,543
b) There are 7 cards numbered 4 each of the 7 rainbow colours. Apart from those we have 42 cards from which we have to choose a hand of 7 thus ensuring that there are no cards numbered 4. So we have to select 7 cards out of 42 which can be done in 42C7 = (42!) / 7!35! = 2,69,78,328
c) There are 3 even numbered cards of each colour:(2,4,6) So a total of 21 even numbered cards and rest 28 odd numbered cards. We can choose 4 even numbered cards out of 21 of them in 21C4 = (21!)/(4!17!) = 5,985 ways
And we can choose 3 odd numbered cards out of 28 of them in 28C3 = (28!)/(3!25!) = 3,276 ways
So number of hands with 4 even and 3 odd numbers = 5985 x 3276 = 1,96,06,860
d) Consider that we have two particular colors of which we want the hand of 7 cards to be made of. For the two colors we get a total of 14 cards. We can select 7 out of these 14 cards in 14C7 = 3,432 ways.
But out of these 3432 ways there are two ways which are not desirable. That is when all seven cards are of either one of the two colors. Because then the criteria that the hand is made of exactly two colors will be violated. Hence we remove those two cases so 3430 ways for any two particular colors.
Now out of the 7 colors we can choose any two in 7C2 = 21 ways
So total number of hands made up of exactly two colors = 21 x 3430 = 72,030
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