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# Combination question - Bridge hands containing 3 aces or 7 diamonds or both?

How many bridge hands contain exactly 3 aces or exactly seven diamonds or both?

One answer is:

4C3 * 48C10 + 13C7 * 39C6 - (12C7 * 3C3 * 36C3 + 1C1 * 12C6 * 3C2 * 36C4)

I know it is the inclusion - exclusion principle. I don't understand the calculation inside the brackets.

hope someone could post an explanation. Thanks.

### 1 Answer

- LearnerLv 71 decade agoFavorite Answer
My explanations below is just confined to only brackets, as you have understood the other one.

Here what we required the combination of 3 Aces and 7 diamonds with any other 3 cards, since in Bridge hand a person will hold 13 cards.

Choice-1: The person receives - one diamond Ace + 6 any other diamonds

from the remaining 12 + 2 Aces from the remaining 3 Aces + any 3 cards

from the remaining 36 cards

NOTE: Here one diamond Ace is the intersection of 4 Aces set and 13 diamond card set; hence the selection one diamond Ace with any 6 other diamonds and other 2 Aces, clearly meets the specification of 3 Aces as well 7 diamonds.

=> the combination = C(1,1) x C(12, 6) x C(3, 2) x C(36,3)

Choice-2: The person receives - Excluding Ace diamond the person receives any 7 from the remaining 12 diamond cards + all 3 Aces from the remaining 3 Aces, leaving diamond Ace + Any 3 cards from the remaining 36 cards (4Aces and 12 diamonds are removed)

=> the combination = C(12, 7) x C(3, 3) x C(36, 3)

Wish you are satisfied with the explanation; have a nice time

Source(s): Self knowledge

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