A seven-digit number comprises of only 2s and 3s. How many of these ar...
Number should be a multiple of 3 and 4. So, the sum of the digits should be a multiple of 3. We can either have all seven digits as 3, or have three 2's and four 3's, or six 2's and a 3.
(The number of 2's should be a multiple of 3).
For the number to be a multiple of 4, the last 2 digits should be 32. Now, let us combine these two.
All seven 3's - No possibility.
Three 2's and four 3's - The first 5 digits should have two 2's and three 3's in some order.
No of possibilities = 5!3!2!5!3!2! = 10
Six 2's and one 3 - The first 5 digits should all be 2's. So, there is only one number 2222232.
So, there are a total of 10 + 1 = 11 solutions.
Hence the answer is "11"
Choice D is the correct answer.
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A seven-digit number comprises of only 2s and 3s. How many of these ar...
Value of 100P2:
The value of 100P2 can be calculated using the formula for permutation. Permutation is the arrangement of objects in a specific order. The formula for permutation is given by:
nP r = n! / (n-r)!
Where n is the total number of objects and r is the number of objects to be arranged in a specific order.
Substituting the values of n and r, we get:
100P2 = 100! / (100-2)!
Simplifying the expression, we get:
100P2 = (100 x 99 x 98 x ... x 1) / (98 x 97 x ... x 1)
100P2 = 100 x 99
100P2 = 9900
Therefore, the value of 100P2 is 9900.
A seven-digit number comprises of only 2s and 3s. How many of these ar...
B/z nCr = n!/(n-r)! So 100P2=100!/(100-2)! =100×99×98!/(98)! =9900