schoolAccedmy
9th10th
11th12th
school

11TH · MATH

Question 4

menu_bookSolutionmenu_bookTheorymenu_book•(i)nCr+nCr−1=n+1Cr{}^nC_r + {}^nC_{r-1} = {}^{n+1}C_rnCr​+nCr−1​=n+1Cr​menu_book•(ii)r⋅nCr=(n−r+1)⋅nCr−1r \cdot {}^nC_r = (n-r+1) \cdot {}^nC_{r-1}r⋅nCr​=(n−r+1)⋅nCr−1​
arrow_backBack
school

11TH · MATH

Question 4

menu_bookSolutionmenu_bookTheorymenu_book•(i)nCr+nCr−1=n+1Cr{}^nC_r + {}^nC_{r-1} = {}^{n+1}C_rnCr​+nCr−1​=n+1Cr​menu_book•(ii)r⋅nCr=(n−r+1)⋅nCr−1r \cdot {}^nC_r = (n-r+1) \cdot {}^nC_{r-1}r⋅nCr​=(n−r+1)⋅nCr−1​
arrow_backBack
Accedmychevron_right11thchevron_rightmathchevron_rightPermutations And Combinationschevron_rightExercise 7.4

See detailed working.

Answer: (computed)

arrow_back
Previous Question
Question 3
Next Question
Question 5
arrow_forward
On this page
No headings yet