Please explain the marked step for this question: $F i n d r, i f 5 {}^{4}P_{r} = 6 {}^{5}P_{r - 1}$
Marked step: $\frac{5!}{(4 - r)!} = \frac{6 \times 5!}{(5 - r + 1) (5 - r) (5 - r - 1)!}$

Question

Please explain the marked step for this question: F i n d   r
,   i f   5 P r 4 = 6 P r - 1 5
Marked step: 5 ! ( 4 - r ) ! = 6 × 5 ! ( 5 - r + 1 ) ( 5 - r
) ( 5 - r - 1 ) !

Ankita Agarwal · Accepted Answer

Dear Student,

5×4!4-r! = 6×5!5-r+1!5! = 5×4×3×2×1  or 5! = 5×4!So, replacing 5×4! by 5! and 6×5! by 6!, we get5!4-r! = 6×5!5-r+1!           
1Now, n! = n×n-1×n-2×n-3!So, 5-r+1!= 5-r+1×5-r+1-1×5-r+1-2!5-r+1!=5-r+15-r5-r-1!Substituting in equ
15!4-r! = 6×5!5-r+15-r5-r-1!14-r!=66-r5-r4-r!By cross multiplication,6-r5-r = 6
Hope this information will clear your doubts about topic

Regards

Please explain the marked step for this question: F i n d r , i f 5 P r 4 = 6 P r - 1 5 Marked step: 5 ! ( 4 - r ) ! = 6 × 5 ! ( 5 - r + 1 ) ( 5 - r ) ( 5 - r - 1 ) !

Please explain the marked step for this question: $F i n d r, i f 5 {}^{4}P_{r} = 6 {}^{5}P_{r - 1}$
Marked step: $\frac{5!}{(4 - r)!} = \frac{6 \times 5!}{(5 - r + 1) (5 - r) (5 - r - 1)!}$