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Parents are having twin babies and have put six unique names card To calculate the number of possible combinations of names for the two babies, you would use the formula for permutations without repetition, which is: nPr = n! (n - r)! Where: So, in this case, the calculation would be: 6P2 = 6! (6 - 2)! This simplifies to: So, there are 30 possible combinations of names for the two babies
Solved Parents are having twin babies and have put 10 unique - Chegg * When the first baby is born, a parent draws a card and gives the baby that name * When the second baby is born, the parent draws a card from the remaining names How many combinations of names were possible for those two babies? Your solution’s ready to go!
Parents are having twin babies and have put 7 unique name cards in a bowl. There are 7 unique name cards in the bowl for the twin babies When the first baby is born, a card is drawn from the bowl and that name is given to the baby When the second baby is born, there is one less card in the bowl, so a card is drawn again
Parents are having twin babies and have put 9 unique name cards in a . . . * When the first baby is born, a parent draws a card and gives the baby that name * When the second baby is born, the parent draws a card from the remaining names How many combinations of names were possible for those two babies?
Parents are having twin babies and have put seven unique name There are 7 unique name cards in the bowl So, there are 7 possible ways to draw a name for the first baby After the first name has been drawn, there are now 6 unique name cards left in the bowl So, there are 6 possible ways to draw a name for the second baby
[Solved] Parents are having twin babies and have put 8 unique name . . . Parents are having twin babies and have put 8 unique name cards in a bowl When the first baby is born, a parent draws a card and gives the baby that name When the second baby is born, the parent draws a card from the remaining names 8 x 7 = 56